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;
}