Category: 解题报告

相关链接

题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1137

解题报告

不难发现每一个点应尽量向编号大的点连边
所以就只剩下$O(n)$条线段了，直接做半平面交就好

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;
 
const int N = 100009;
const double EPS = 1e-6;
 
int n, m, tot;
vector<int> del[N];
struct Point{
    double x, y;
    inline Point() {
    }
    inline Point(double _x, double _y):x(_x), y(_y) {
    }
    inline Point operator - (const Point &P) const {
        return Point(x - P.x, y - P.y);
    }
    inline Point operator + (const Point &P) {
        return Point(x + P.x, y + P.y);
    }
    inline Point operator * (double d) {
        return Point(x * d, y * d);
    }
    inline double operator * (const Point &P) {
        return x * P.y - y * P.x;
    }
    inline double len() {
        return sqrt(x * x + y * y);
    }
}p[N];
struct Line{
    Point a, b;
    double slope;
    inline Line() {
    }
    inline Line(const Point &_a, const Point &_b):a(_a), b(_b) {
    }
    inline bool operator < (const Line &L) const {
        return slope < L.slope;
    }
    inline Point intersect(const Line &L) {
        double s1 = (L.b - L.a) * (b - L.a);
        double s2 = (a - L.a) * (L.b - L.a);
        return a + (b - a) * (s2 / (s1 + s2));
    }
    inline bool OnLeft(const Point &P) {
        return (b - a) * (P - a) > EPS;
    }
}l[N];
 
inline int read() {
    char c = getchar(); int ret = 0, f = 1;
    for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
    for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
    return ret * f;
}
 
inline void HalfPlaneIntersection(Line *arr, int &sz) {
    for (int i = 1; i <= sz; i++) {
        Point vec = arr[i].b - arr[i].a;
        arr[i].slope = atan2(vec.y, vec.x);
    }
    sort(arr + 1, arr + 1 + sz);
    int l = 1, r = 0;
    for (int i = 1; i <= sz; i++) {
        for (; l < r && !arr[i].OnLeft(arr[r].intersect(arr[r - 1])); r--); 
        for (; l < r && !arr[i].OnLeft(arr[l].intersect(arr[l + 1])); l++);
        arr[++r] = arr[i];
    }
    for (; l < r - 1 && !arr[l].OnLeft(arr[r].intersect(arr[r - 1])); r--);
    sz = 0;
    for (int i = l; i <= r; i++) {
        arr[++sz] = arr[i];
    }
    arr[0] = arr[sz];
    arr[sz + 1] = arr[1];
}
 
int main() {
    n = read(); m = read();
    for (int i = 1; i <= n; i++) {
        p[i].x = read();
        p[i].y = read();
    }
    for (int i = 1; i <= m; i++) {
        int x = read(), y = read();
        del[min(x, y)].push_back(max(x, y));
    }
    for (int i = 1; i < n; i++) {
        sort(del[i].begin(), del[i].end());
        int to = n;
        while (!del[i].empty() && *--del[i].end() == to) {
            to--;
            del[i].pop_back();
        }
        if (i == 1 && to == n) {
            printf("%.10lf", (p[1] - p[n]).len());
            exit(0);
        }
        if (to > i) {
            l[++tot] = Line(p[to], p[i]);
        }
    }
    l[++tot] = Line(p[1], p[n]); 
    HalfPlaneIntersection(l, tot);
    double ans = -(p[1] - p[n]).len();
    Point last = l[1].intersect(l[0]);
    for (int i = 1; i <= tot; i++) {
        Point cur = l[i].intersect(l[i + 1]);
        ans += (cur - last).len();
        last = cur;
    }
    printf("%.10lf\n", ans);
    return 0;
}

相关链接

题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4530

解题报告

这题LCT是可以做的

但因为本题没有要求强制在线
所以我们可以用并查集来维护连通性，再用线段树合并来支持支持DFS序的区间询问
总的时间复杂度：$O(n \log n)$

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 200009; 
const int M = N * 21;

int n, m, vis[N], head[N], nxt[N], to[N];
int dep[N], beg[N], out[N], sz[N], fa[N];
struct Data{
	int t, x, y;
	inline Data() {
	}
	inline Data(bool a, int b, int c):t(a), x(b), y(c) {
	} 
}opt[N];

inline int read() {
	char c = getchar(); int ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}

inline void AddEdge(int u, int v) {
	static int E = 1;
	to[++E] = v; nxt[E] = head[u]; head[u] = E;
	to[++E] = u; nxt[E] = head[v]; head[v] = E;
}

inline void DFS(int w, int f) {
	static int D = 0;
	vis[w] = 1;
	beg[w] = ++D;
	dep[w] = dep[f] + 1;
	for (int i = head[w]; i; i = nxt[i]) {
		if (to[i] != f) {
			DFS(to[i], w);
		}
	}
	out[w] = D;
}

inline int find(int x) {
	return fa[x] == x? x: fa[x] = find(fa[x]);
}

class SegmentTree{
int cnt, ch[M][2], sum[M], root[N];
public:
	inline void insert(int p, int v) {
		insert(root[p], 1, n, v);
	}
	inline void merge(int a, int b) {
		root[a] = Merge(root[a], root[b]);
	}
	inline int query(int p, int l, int r) {
		return query(root[p], 1, n, l, r);
	}
private:
	inline int Merge(int a, int b) {
		if (!a || !b) {
			return a + b;
		} else {
			sum[a] += sum[b];
			ch[a][0] = Merge(ch[a][0], ch[b][0]);
			ch[a][1] = Merge(ch[a][1], ch[b][1]);
			return a;
		}
	}
	inline void insert(int &w, int l, int r, int p) {
		sum[w = ++cnt] = 1;
		if (l < r) {
			int mid = l + r + 1 >> 1;
			if (p < mid) {
				insert(ch[w][0], l, mid - 1, p);
			} else {
				insert(ch[w][1], mid, r, p);
			}
		}
	} 
	inline int query(int w, int l, int r, int L, int R) {
		if (!w) {
			return 0;
		} else if (L <= l && r <= R) {
			return sum[w];
		} else {
			int mid = l + r + 1 >> 1, ret = 0;
			ret += L < mid? query(ch[w][0], l, mid - 1, L, R): 0;
			ret += mid <= R? query(ch[w][1], mid, r, L, R): 0;
			return ret;
		}
	}
}SGT;

int main() {
	n = read(); m = read();
	for (int i = 1; i <= m; i++) {
		char cmd[3]; 
		scanf("%s", cmd);
		int u = read(), v = read();
		if (cmd[0] == 'A') {
			AddEdge(u, v);
		}
		opt[i] = Data(cmd[0] == 'A', u, v);
	}
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {
			DFS(i, i);
		}
	}
	for (int i = 1; i <= n; i++) {
		sz[i] = 1;
		fa[i] = i;
		SGT.insert(i, beg[i]);
	}
	for (int i = 1; i <= m; i++) {
		int u = opt[i].x, v = opt[i].y;
		if (opt[i].t == 1) {
			SGT.merge(find(u), find(v));
			sz[find(u)] += sz[find(v)];
			fa[find(v)] = find(u);
		} else {
			if (dep[u] < dep[v]) {
				swap(u, v);
			}
			int p1 = SGT.query(find(u), beg[u], out[u]);
			int p2 = sz[find(u)] - p1;
			printf("%lld\n", (LL)p1 * p2);
		}
	}
	return 0;
}

相关链接

题目传送门：http://oi.cyo.ng/wp-content/uploads/2017/07/20170624_problem.pdf

解题报告

我们往后多看一步：

假设下一步我们需要向左走，那么我们这一步就走到对于现在所在的点极角序最小的点去
否则就走到极角序最大的点去

不难发现这样一定能构造出一组解

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 5009;

int n, vis[N];
char cmd[N];
struct Point{
	int x, y;
	inline Point() {
	}
	inline Point(int a, int b):x(a), y(b) {
	}
	inline Point operator - (const Point &P) {
		return Point(x - P.x, y - P.y);
	}
	inline bool operator < (const Point &P) {
		return x < P.x || (x == P.x && y < P.y);
	}
	inline LL operator * (const Point &P) {
		return x * P.y - y * P.x;
	}
}p[N];

inline int read() {
	char c = getchar(); int ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}

inline void solve(int w, int stp) {
	vis[w] = 1;
	printf("%d ", w);
	if (stp < n) {
		int nxt = 0;
		for (int i = 1; i <= n; i++) {
			if (!vis[i]) {
				if (!nxt || (cmd[stp] == 'L'? (p[nxt] - p[w]) * (p[i] - p[w]) < 0: (p[nxt] - p[w]) * (p[i] - p[w]) > 0)) {
					nxt = i;
				}
			}
		}
		solve(nxt, stp + 1);
	}
}

int main() {
	freopen("route.in", "r", stdin);
	freopen("route.out", "w", stdout);
	n = read();
	int s = 0;
	for (int i = 1; i <= n; i++) {
		p[i].x = read();
		p[i].y = read();
		if (!s || p[i] < p[s]) {
			s = i;
		}
	}
	scanf("%s", cmd);
	solve(s, 1);
	return 0;
}

相关链接

题目传送门：http://agc015.contest.atcoder.jp/tasks/agc015_f

解题报告

我们写出使答案最大且值最小的序列，不难发现这是一个斐波那契数列
进一步发现，这条主链的每一个点只会引申出一条支链，且支链不会再分
所以我们可以暴力算出了$\log n$条链，共$\log ^ 2 n$对数，然后暴力算贡献

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;
 
const int MOD = 1000000007;
const int LOG = 100;
 
vector<pair<LL, LL> > num[LOG];
 
inline LL read() {
	char c = getchar(); LL ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}
 
inline void solve(LL A, LL B) {
	if (A < num[2][0].first || B < num[2][0].second) {
		printf("1 %d\n", A * B % MOD);
		return;
	}
	for (int i = 2; i < LOG; i++) {
		if (A < num[i + 1][0].first || B < num[i + 1][0].second) {
			int cnt = 0;
			for (int j = 0; j < (int)num[i].size(); j++) {
				LL a = num[i][j].first, b = num[i][j].second;
				cnt = (cnt + (a <= A && b <= B? (B - b) / a + 1: 0)) % MOD;
				cnt = (cnt + (b <= A && a <= B? (A - b) / a + 1: 0)) % MOD;
			}
			printf("%d %d\n", i, cnt);
			return;
		}
	}
}
 
inline void prework() {
	num[0] = vector<pair<LL,LL> >{make_pair(1, 1)};
	for (int i = 1; i < LOG; i++) {
		for (int j = 0; j < (int)num[i - 1].size(); ++j) {
			LL a = num[i - 1][j].first, b = num[i - 1][j].second;
			num[i].push_back(make_pair(b, a + b));
		}
		LL a = num[i - 1][0].first, b = num[i - 1][0].second;
		num[i].push_back(make_pair(a + b, 2 * a + b));
	}
}
 
int main() {
	prework();
	for (int T = read(); T; T--) {
		LL A = read(), B = read();
		solve(min(A, B), max(A, B));
	}
	return 0;
}

相关链接

题目传送门：http://oi.cyo.ng/wp-content/uploads/2017/07/NOI2017-matthew99.pdf

解题报告

我们按照极角序来贪心匹配
不难发现这样一定有解

另外Dinic是不能求这种要求字典序最小的解的
似乎只能用最裸的匈牙利

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 1009;
const int M = 1000009;
const int INF = 1e9;
const double PI = acos(-1);

int n, R, D, S, T;
int in[N], ot[N], cnt_in, cnt_ot;
int head[N], nxt[M], to[M], mth[N], vis[N];
pair<double, int> arr[N];
struct Point{
	int x, y, ty;
	inline int dis(const Point &P) {
		return (x - P.x) * (x - P.x) + (y - P.y) * (y - P.y);
	}
}p[N];

inline void AddEdge(int u, int v) {
	static int E = 1;
	to[++E] = v; nxt[E] = head[u]; head[u] = E;
} 

inline int read() {
	char c = getchar();
	int ret = 0, f = 1;
	while (c < '0' || c > '9') {
		f = c == '-'? -1: 1;
		c = getchar();
	}
	while ('0' <= c && c <= '9') {
		ret = ret * 10 + c - '0';
		c = getchar();
	}
	return ret * f;
}

inline bool DFS(int w) {
	vis[w] = 1;
	for (int i = head[w]; i; i = nxt[i]) {
		if (!mth[to[i]] || (!vis[mth[to[i]]] && DFS(mth[to[i]]))) {
			mth[to[i]] = w;
			mth[w] = to[i];
			return 1;
		} 
	}
	return 0;
}

inline int solve() {
	int ret = 0;
	for (int ii = 1, i; i = in[ii], ii <= cnt_in; ii++) {
		if (!mth[i]) {
			memset(vis, 0, sizeof(vis));
			ret += DFS(i);
		}
	}
	return ret;
}

inline void print(int w) {
	for (int i = head[w]; i; i = nxt[i]) {
		if (to[i] == mth[w]) {
			vis[w] = vis[to[i]] = 1;
			printf("%d %d\n", w, mth[w]);
			return;
		} else if (!vis[to[i]]){
			print(mth[to[i]]);
		}
	}
}

int main() {
	n = read(); 
	R = read(); R *= R;
	D = read(); D *= D;
	S = 0; T = N - 1;
	for (int i = 1; i <= n; i++) {
		p[i].x = read();
		p[i].y = read();
		if (p[i].ty = p[i].dis(p[1]) > R) {
			in[++cnt_in] = i;
		} else {
			ot[++cnt_ot] = i;
		}
	}
	for (int ii = 1; ii <= cnt_in; ii++) {
		int i = in[ii], cnt_arr = 0;
		double mx = -PI, mn = PI;
		for (int jj = 1; jj <= cnt_ot; jj++) {
			int j = ot[jj];
			if (p[i].dis(p[j]) <= D) {
				double agl = atan2(p[j].y - p[i].y, p[j].x - p[i].x);
				mx = max(mx, agl);
				mn = min(mn, agl);
				arr[++cnt_arr] = make_pair(agl, j);
			}
		}
		if (mx - mn >= PI) {
			for (int j = 1; j <= cnt_arr; j++) {
				arr[j].first += arr[j].first < 0? PI * 2: 0;
			}
		}
		sort(arr + 1, arr + 1 + cnt_arr);
		for (int j = 1; j <= cnt_arr; j++) {
			AddEdge(i, arr[j].second);
		}
	}
	printf("%d\n", solve() << 1);
	memset(vis, 0, sizeof(vis));
	for (int ii = 1, i; i = in[ii], ii <= cnt_in; ii++) {
		if (mth[i] && !vis[i]) {
			print(i);
		}
	}
	return 0;
}

相关链接

题目传送门：http://oi.cyo.ng/wp-content/uploads/2017/07/20170623_statement.pdf

解题报告

假如我们钦定某个格子先取走靠左的三角形，那么其余格子是先取走靠左还是靠右就全部定下来了
于是我们暴力枚举一下，复杂度是$O(n^4)$

更进一步，我们发现：

假如我们钦定先取走$(x, y)$这个格子的靠左三角形
那么我们一定得先将$(x – 1, y)$这个格子的左三角形取走，然后再取走一些其他的三角形

于是同一行的信息是可以共用的，然后就可以记忆化搜索了
时间复杂度是$O(n^3)$的

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 500;
const int INF = 1e7;

char mp[N][N];
int n, m, ans[N];
bool up[N][N], dw[N][N], InStack[N][N], vis[N][N];

inline int read() {
	char c = getchar(); int ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}

inline int F(int x, int y, int t) {
	InStack[y][x] = 1;
	int nx = x + (t == 1? 1: -1), ny = y, nt = t, ret = 1;
	ret += vis[ny][nx]? 0: (InStack[ny][nx]? INF: F(nx, y, t));
	nx = x; ny = up[y][x] == t? y - 1: y + 1; nt = up[y][x] == t? up[ny][nx]: dw[ny][nx];
	ret += vis[ny][nx] || ret >= INF? 0: (InStack[ny][nx]? INF: F(nx, ny, nt));	
	vis[y][x] = 1;
	InStack[y][x] = 0;
	return ret > INF? INF: ret;
}

inline void init() {
	memset(vis, 0, sizeof(vis));
	for (int j = 1; j <= m; j++) {
		vis[j][0] = 1;
		vis[j][n + 1] = 1;
	}
	for (int i = 1; i <= n; i++) {
		vis[0][i] = 1;
		vis[m + 1][i] = 1;
	}
}

int main() {
	m = read(); n = read();
	for (int j = 1; j <= m; j++) {
		scanf("%s", mp[j] + 1);
		for (int i = 1; i <= n; i++) {
			up[j][i] = 1;
			dw[j][i] = 0;
			if (mp[j][i] == 'Z') {
				swap(up[j][i], dw[j][i]);
			}
		}
	}
	for (int j = 1; j <= m; j++) {
		init();
		for (int i = n; i; i--) {
			ans[i] = ans[i + 1] < INF? F(i, j, 1) + ans[i + 1]: INF;
		}
		init();
		for (int i = 1; i <= n; i++) {
			ans[i] = min(ans[i], ans[i - 1] < INF? F(i, j, 0) + ans[i - 1]: INF);
			printf("%d ", ans[i] >= INF? -1: ans[i] << 1);
		}
		putchar('\n');
	}
	return 0;
}

相关链接

题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2471
神犇题解：http://www.cnblogs.com/clrs97/p/5993606.html

解题报告

我们考虑从高位开始，逐位枚举。那么每枚举一位相当于将序列划分为10分，并且取其中一份。
对于一段连续的序列来讲，我们只需要关注其进入这段序列之前会匹配到x的哪一位、匹配完这一段之后匹配到了x的哪一位、这期间总共贡献了多少次成功的匹配。
不难发现这个状态是很少的，于是我们可以记忆化搜索。

另外这题很容易扩展到：“左右边界为任意值的情况”
然后我把这题搬到了今年的全国胡策里，不知道有没有人能切掉

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;
 
const int N = 16;
const int M = 10;
const int SGZ = 10;
const int MOD = 1000000007;
 
int n, m, nxt[M];
char s[M];
struct Transfer{
    LL pos, cnt;
    inline Transfer() {
    }
    inline Transfer(LL a, LL b):pos(a), cnt(b) {
    }
    inline Transfer operator + (const Transfer &T) {
        return Transfer(T.pos, cnt + T.cnt);
    }
}t[M][SGZ];
map<int, Transfer> f[N][M];
struct MatchStatus{
    int HashVal;
    Transfer t[M];
    inline void GetHashVal() {
        const static int MOD = 1000000007;
        const static int SEED1 = 13, SEED2 = 131;
        HashVal = 0;
        for (int i = 0; i < m; i++) {
            HashVal = (HashVal + (LL)t[i].pos * SEED2 + t[i].cnt) * SEED1 % MOD;
        }
    }
    inline bool operator < (const MatchStatus &MS) const {
        return HashVal < MS.HashVal;
    }
};
 
inline Transfer DFS(int w, int p, const MatchStatus &cur) {
    if (w <= 0) {
        return cur.t[p];
    } else if (f[w][p].count(cur.HashVal)) {
        return f[w][p][cur.HashVal];
    } else {
        Transfer ret(p, 0);
        for (int i = 0; i < SGZ; i++) {
            MatchStatus nw = cur;
            for (int j = 0; j < m; j++) {
                nw.t[j] = nw.t[j] + t[nw.t[j].pos][i];
            }
            nw.GetHashVal();
            ret = ret + DFS(w - 1, ret.pos, nw);
        }
        return f[w][p][cur.HashVal] = ret;
    }
}
 
int main() {
    while (~scanf("%d %s", &n, s + 1) && n) {
        m = strlen(s + 1);
        for (int i = 1; i <= m; i++) {
            s[i] -= '0';
        }
        nxt[1] = 0;
        for (int i = 2, j = 0; i <= m; nxt[i++] = j) {
            for (; j && s[j + 1] != s[i]; j = nxt[j]);
            j += s[j + 1] == s[i];
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < SGZ; j++) {
                int k = i;
                for (; k && s[k + 1] != j; k = nxt[k]);
                k += s[k + 1] == j;
                t[i][j] = k == m? Transfer(nxt[k], 1): Transfer(k, 0);
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < m; j++) {
                f[i][j].clear();
            }
        }
        Transfer ans(0, 0);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j < SGZ; j++) {
                MatchStatus cur;
                for (int k = 0; k < m; k++) {
                    cur.t[k] = t[k][j];
                }
                cur.GetHashVal();
                ans = ans + DFS(i - 1, ans.pos, cur);
            }
        }
        printf("%lld\n", ans.cnt);
    }
    return 0;
}