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  • HDU 3879 Base Station(最大权闭合子图)

    将第i个用户和他需要的基站连边,转化成求二分图的最大权闭合子图。

    答案=正权点之和-最小割。

    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi 3.1415926535
    # define eps 1e-9
    # define MOD 1000000009
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int res=0, flag=0;
        char ch;
        if((ch=getchar())=='-') flag=1;
        else if(ch>='0'&&ch<='9') res=ch-'0';
        while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
        return flag?-res:res;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=60005;
    //Code begin...
    
    struct Edge{int p, next, w;}edge[350005];
    int head[N], cnt=2, s, t, vis[N];
    queue<int>Q;
    
    void add_edge(int u, int v, int w){
        edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
        edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++;
    }
    int bfs(){
        int i, v;
        mem(vis,-1);
        vis[s]=0; Q.push(s);
        while (!Q.empty()) {
            v=Q.front(); Q.pop();
            for (i=head[v]; i; i=edge[i].next) {
                if (edge[i].w>0 && vis[edge[i].p]==-1) {
                    vis[edge[i].p]=vis[v] + 1;
                    Q.push(edge[i].p);
                }
            }
        }
        return vis[t]!=-1;
    }
    int dfs(int x, int low){
        int i, a, temp=low;
        if (x==t) return low;
        for (i=head[x]; i; i=edge[i].next) {
            if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){
                a=dfs(edge[i].p,min(edge[i].w,temp));
                temp-=a; edge[i].w-=a; edge[i^1].w += a;
                if (temp==0) break;
            }
        }
        if (temp==low) vis[x]=-1;
        return low-temp;
    }
    int main ()
    {
        int n, m, u, v, tmp;
        while (~scanf("%d%d",&n,&m)) {
            s=0, t=n+m+1;
            int res=0, sum=0;
            mem(head,0); cnt=2;
            FOR(i,1,n) scanf("%d",&tmp), add_edge(m+i,t,tmp);
            FOR(i,1,m) {
                scanf("%d%d%d",&u,&v,&tmp); res+=tmp;
                add_edge(s,i,tmp); add_edge(i,m+u,INF); add_edge(i,m+v,INF);
            }
            while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp;
            printf("%d
    ",res-sum);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6635287.html
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