Category: 网络流

题目传送门：http://cogs.top/cogs/problem/problem.php?pid=396
离线版题目：http://paste.ubuntu.com/18780019/

贪心即可。因为枚举一下，1600以内，加上比自己小的数为完全平方数没有多解，所以只要能加，加上去就好
当然， 正解是最少路径覆盖+二分，然而为什么要闲得蛋疼写Dinic ╮(╯-╰)╭
另外，COGS的浮点运算取整是辣鸡QAQ，害的我wa了好几发……

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

const int MAXN = 10000;
const int MX = 1600;

int n,tot,arr[MAXN],vout;

inline bool judge(int num){
	for (int i=1;i<=tot;i++){
		int tmp = sqrt(arr[i]+num);
		if (tmp*tmp == arr[i]+num){
			arr[i] = num; 
			return true;
		}
	}
	if (tot < n) {arr[++tot] = num; return true;}
	return false;
}

int main(){  
	freopen("balla.in","r",stdin);
	freopen("balla.out","w",stdout);
	scanf("%d",&n);
	for (int i=1;i<=MX;i++)
		if (judge(i)) vout=i;
		else {printf("%d\n",i-1);break;}
	return 0;
} 

题目传送门：http://cogs.top/cogs/problem/problem.php?pid=14
离线版题目：http://paste.ubuntu.com/18772859/

二分图匹配板题，Dinic跑一跑就ok啦！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

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

const int MAXN = 10000+9;
const int INF = 1000000000;

int n,m,a,b,T,head[MAXN],to[MAXN],nxt[MAXN],cap[MAXN],vout;
int s,t,cur[MAXN],que[MAXN],fro,bak,dis[MAXN],flow[MAXN];

inline void AddEdge(int a, int b){
	to[++T] = b; nxt[T] = head[a]; head[a] = T; cap[T] = 1;
	to[++T] = a; nxt[T] = head[b]; head[b] = T;  
}

inline bool BFS(){
	memset(dis,-1,sizeof(dis));
	dis[s] = 0; que[fro=bak=1]=s;
	
	while (bak <= fro){
		int w = que[bak++];
		for (int i=head[w];i;i=nxt[i])
			if (flow[i] < cap[i] && dis[to[i]]==-1)
				dis[to[i]] = dis[w] + 1,
				que[++fro] = to[i];
	}
	
	if (dis[t] != -1) return true;
	else return false;
}

int MaxFlow(int w, int f){
	if (w == t) return f;
	else {
		int ret = 0;
		for (int &i=cur[w];i;i=nxt[i]){
			if (flow[i] < cap[i] && dis[to[i]] == dis[w]+1){
				int tmp = MaxFlow(to[i], min(f, cap[i]-flow[i]));
				flow[i] += tmp; flow[i^1] -= tmp;
				ret += tmp; f-=tmp;
				if (!f) return ret;
			}
		}
		return ret;
	}
}

inline void Dinic(){
	while (BFS()){
		for (int i=0;i<=n*2+1;i++)
			cur[i] = head[i];
		vout += MaxFlow(s,INF);
	}
}

int main(){
	freopen("flyer.in","r",stdin);
	freopen("flyer.out","w",stdout);
	n = read(); m = read();
	s = 0; t = 2*n+1; T = 1;
	while (~scanf("%d%d",&a,&b)) AddEdge(a*2,b*2-1);
	for (int i=1;i<=m;i++) AddEdge(s,i*2-1);
	for (int i=m+1;i<=n;i++) AddEdge(i*2,t);
	for (int i=1;i<=n;i++) AddEdge(i*2-1,i*2);
	Dinic();
	printf("%d\n",vout);
	return 0;
}

题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1001
数据生成器：http://paste.ubuntu.com/18772741/

最大流板题，输入搞反了，调了3hQAQ

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
 
const int N = 1000000+9;
const int INF = 1000000000;
 
inline int read(){
    char c=getchar(); int buf=0,f=1;
    while (c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while (c<='9'&c>='0'){buf=buf*10+c-'0';c=getchar();}
    return buf*f;
}
 
int n,m,head[N],to[N*6],nxt[N*6],cap[N*6],T,flow[N*6];
int vout,cur[N],que[N],dis[N],fro,bak,tot,s,t;
 
#define id(x,y) (((y)-1)*n+(x))
 
inline void AddEdge(int u, int v, int MX){
    to[++T] = v; nxt[T] = head[u]; head[u] = T; cap[T] = MX;
    to[++T] = u; nxt[T] = head[v]; head[v] = T; cap[T] = MX;
}
 
inline bool BFS(){
    memset(dis,-1,sizeof(dis));
    que[fro=bak=1] = s; dis[s] = 0;
     
    while (bak <= fro) {
        int w = que[bak++];
        for (int i=head[w];i;i=nxt[i]){
            if (flow[i] < cap[i] && dis[to[i]] == -1)
                dis[to[i]] = dis[w]+1,
                que[++fro] = to[i];
        }
    }
    if (dis[t] == -1) return false;
    else return true;
}
 
int MaxFlow(int w, int f){
    if (w==t) {vout += f;return f;}
    else {
        int ret = 0;
        for (int &i=cur[w],tmp;i;i=nxt[i]){
            if (flow[i] < cap[i] && dis[to[i]]==dis[w]+1){
                tmp = MaxFlow(to[i], min(f, cap[i]-flow[i]));
                f -= tmp; ret += tmp;
                flow[i] += tmp; flow[i^1] -= tmp;
                if (!f) return ret;
            }
        }
        return ret;
    }
}
 
inline void Dinic(){
    tot = id(n,m);
    while (BFS()){
        for (int i=1;i<=tot;i++)
            cur[i] = head[i];
        MaxFlow(s,INF);
    }   
}
 
int main(){
    m = read(); n = read(); s = id(1,1); t = id(n,m); T = 1;
    for (int j=1;j<=m;j++) for (int i=1;i<n;i++) AddEdge(id(i,j),id(i+1,j), read());
    for (int j=1;j<m;j++) for (int i=1;i<=n;i++) AddEdge(id(i,j),id(i,j+1), read());
    for (int j=1;j<m;j++) for (int i=1;i<n;i++) AddEdge(id(i,j),id(i+1,j+1),read());
    if (s != t) Dinic();
    printf("%d\n",vout);
    return 0;
}

最近几天，重新来看网络流的相关题目，发现之前看的基本上都忘了 QAQ
看了一个上午，有以下收获
1.最大流的正确性依赖于每一条s-t流对应了一种实际方案
2.网络流也可以作为二分的可行解判定方式
3.网络流只能在0/1之中做选择，对于1/2、2/3之类的，可以考虑降到0/1之后算贡献
4.对于割边，可以在残余网络中DFS搞一搞，一端可到源点，一端可到汇点即为割边

另外，提供一点网络流24题的相关资源：
1.完整题解：http://www.snowyjone.me/2015/07/lp-and-flow-a/
2.精选题解：https://blog.sengxian.com/solutions/networkflow-24-all
3.大神题解：https://www.byvoid.com/blog/lpf24-solution
4.OJ传送门：http://cojs.tk/cogs/page/page.php?aid=3