（网络流）LGP2691

luoguP2691
n*n 的网格图中要从m个点中找互不相交的路径
我们考虑如何建图
首先将起点S连向ｍ个点
将图的边缘连向终点Ｔ
为使每个点只经过一次将每个点拆分为两个点
让边权（流量）为一跑dinic

#include<bits/stdc++.h>
using namespace std;
const int N = 10100;
int n,m;
int tot,head[N],to[N<<2],val[N<<2],nex[N<<2],op[N<<2];
void add(int x,int y,int f_){
 //   cout << "Debug " << x << ' ' << y  << endl;
    to[++tot] = y;val[tot] = f_;nex[tot] = head[x];head[x] = tot;
    to[++tot] = x;val[tot] = 0;nex[tot] = head[y];head[y] = tot;
    op[tot] = tot-1;op[tot-1] = tot;
}
int dx[5] = {0,1,-1,0,0};
int dy[5] = {0,0,0,1,-1};
int s,t;
int dep[10010];
bool bfs(){
    for(int i = 0;i <= n * n * 2 + 1;i ++){
        dep[i] = 0;
    }
    queue<int>q;
    dep[s] = 1;
    q.push(s);
    while(q.size()){
        int x = q.front();
        q.pop();
        for(int i = head[x];i;i = nex[i]){
            int y = to[i];int f_ = val[i];
            if(f_==0 || dep[y])continue;
            dep[y] = dep[x] + 1;
            q.push(y);
        }
    }
    return dep[t] != 0;
}
int dfs(int x,int flow){
    if(x == t)return flow;
    int ans = 0;
    for(int i = head[x];i;i = nex[i]){
        int y = to[i];int f_ = val[i];
        if(f_==0 || dep[y] != dep[x] + 1||flow == 0)continue;
        int detal = dfs(y,min(f_,flow));
        if(detal == 0)dep[y] = -1;
        ans += detal;flow -= detal;
        val[i] -= detal;val[op[i]] += detal;
    }
    return ans;
}
int dinic(){
    int ans = 0;
    while(bfs())ans += dfs(s,0x3f3f3f3f);
    return ans;
}

int main(){
    scanf("%d%d",&n,&m);
    s = 0;t = n * n * 2 + 1;
    for(int i = 1;i <= n;i ++)
        for(int j = 1;j <= n;j ++){
            add((i-1)*n+j ,n*n+(i-1)*n+j,1);// enter -> away
            for(int k = 1;k <= 4;k ++){
                if(i + dx[k] > n || i + dx[k] < 1 || j + dy[k] > n || j + dy[k] < 1)continue;
                int x = i + dx[k];int y = j + dy[k];
                add(n*n+(i-1)*n+j , (x-1)*n+y, 1);
            }
        }
    
    for(int i = 1;i <= n;i ++){add(n*n+i,t,1);add(n*n+(n-1)*n+i,t,1);}
    for(int i = 2;i <= n-1;i ++){add(n*n+(i-1)*n+1,t,1);add(n*n+(i-1)*n+n,t,1);}
    for(int i = 1;i <= m;i ++){
        int x,y;scanf("%d%d",&x,&y);
        add(s,(x-1)*n+y,1);
    }
    printf(dinic()>=m ? "YES" : "NO");
    return 0;
}