Codeforces 1082G(最大权闭合子图）

题面

传送门

分析

没想到压轴题是道模板裸题

由于子图的权值=边权和-点权和

将边和点都看成新图中的点

S向原来的边i连边,权值为边权

点i向T连边，权值为点权

边i:(u,v,w)向u,v,连边，权值INF

答案即为总边权-最小割

理论时间复杂度为\(O(n^2m)\)，但由于Dinic在随机图上表现很好，可以通过

代码

#include<iostream>
#include<cstdio>
#include<cstring> 
#include<queue>
#define maxn 10005
#define maxm 10005
#define INF 0x3f3f3f3f3f3f3f3f 
using namespace std;
int n,m;
struct edge{
    int from;
    int to;
    int next;
    long long flow;
}E[maxm<<1];
int sz=1;
int head[maxn];
void add_edge(int u,int v,long long w){
//  printf("%d->%d : %d\n",u,v,w);
    sz++;
    E[sz].from=u;
    E[sz].to=v;
    E[sz].next=head[u];
    E[sz].flow=w;
    head[u]=sz;
    sz++;
    E[sz].from=v;
    E[sz].to=u;
    E[sz].next=head[v];
    E[sz].flow=0;
    head[v]=sz;
}

int deep[maxn];
bool bfs(int s,int t){
    queue<int>q;
    memset(deep,0,sizeof(deep));
    q.push(s);
    deep[s]=1;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=head[x];i;i=E[i].next){
            int y=E[i].to;
            if(E[i].flow&&!deep[y]){
                deep[y]=deep[x]+1;
                if(y==t) return 1;
                q.push(y);
            } 
        }
    }
    return 0;
}

long long dfs(int x,int t,long long minf){
    if(x==t) return minf;
    long long k,rest=minf;
    for(int i=head[x];i;i=E[i].next){
        int y=E[i].to;
        if(E[i].flow&&deep[y]==deep[x]+1){
            k=dfs(y,t,min(rest,E[i].flow));
            if(k==0) deep[y]=0;
            E[i].flow-=k;
            E[i^1].flow+=k;
            rest-=k; 
            if(rest==0) break;
        }
    }
    return minf-rest;
}

long long dinic(int s,int t){
    long long maxflow=0,nowflow=0;
    while(bfs(s,t)){
        while(nowflow=dfs(s,t,INF)) maxflow+=nowflow;
    }
    return maxflow;
}

int a[maxn];
int main(){
    int u,v,w;
    scanf("%d %d",&n,&m);
    int s=0,t=n+m+1;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        add_edge(i,t,a[i]);
    }
    long long sum=0;
    for(int i=1;i<=m;i++){
        scanf("%d %d %d",&u,&v,&w);
        sum+=w;
        add_edge(s,i+n,w);
        add_edge(i+n,u,INF);
        add_edge(i+n,v,INF);
    }
    printf("%I64d\n",sum-dinic(s,t));
}