网络流24题9

把棋盘的相邻关系转化为二分图做

然后就是最大独立点集

然后带权的就用网络流

因为每个割对应一种取独立点集的方案

跑一边最小割

https://loj.ac/problem/6007

#include <iostream>
#include <queue>
#include <cstring>
#include <stdio.h>
using namespace std;
const int maxn = 109;
int a[maxn][maxn];
int dir[4][2] = {1,0,-1,0,0,1,0,-1};
int head[maxn*maxn];
int tot;
struct edge{
    int v,nex,w;
}e[maxn*maxn*8];
void addedge(int u,int v,int w){
    e[tot] = (edge){v,head[u],w};
    head[u] = tot++;
    e[tot] = (edge){u,head[v],0};
    head[v] = tot++;
}
int deep[maxn*maxn];
bool bfs(int S,int T){
    memset(deep,0,sizeof(deep));
    deep[S] = 1;
    queue<int> q;
    q.push(S);
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].nex){
            int v = e[i].v;
            int w = e[i].w;
            if(w<=0 || deep[v]!=0) continue;
            deep[v] = deep[now]+1;
            q.push(v);
        }
    }
    return deep[T];
}
int dfs(int now,int T,int maxflow){
    if(now==T) return maxflow;
    int all = 0;
    for(int i=head[now];i!=-1 && all<maxflow;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(w<=0 || deep[v]!=deep[now]+1) continue;
        int tt =dfs(v,T,min(maxflow-all,w));
        e[i].w-=tt;
        e[i^1].w+=tt;
        all+=tt;
    }
    return all;
}
int dinic(int S,int T){
    int ret = 0;
    while(bfs(S,T)){
        ret+=dfs(S,T,0x3f3f3f3f);
    }
    return ret;
}
int main()
{
    int n,m;
    scanf("%d%d",&m,&n);
    memset(head,-1,sizeof(head));
    int T=(n+1)*m+1;
    tot = 0;
    int all = 0;
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            int t;
            scanf("%d",&t);
            all +=t;
            if((i+j)%2==0){
                addedge(0,i*n+j,t);
                for(int k=0;k<4;k++){
                    int nx = i+dir[k][0];
                    int ny = j+dir[k][1];
                    if(nx<=m && nx>=1 && ny<=n&&ny>=1){
                        addedge(i*n+j,nx*n+ny,0x3f3f3f3f);
                    }
                }
                
            }else{
                addedge(i*n+j,T,t);
            }
        }
    }
    int ans = all-dinic(0,T);
    printf("%d\n",ans);
    return 0;
}