bzoj1497 [NOI2006]最大收益

来源：互联网发布：剑三画面优化编辑：程序博客网时间：2024/05/04 11:34

最小割思路比较简单
点数看起来很多但是因为是类似二分图的东西所以跑的比较快
源点对每个用户连容量为收益的边用户向中转站连容量无穷大的边中转站向汇点连容量为成本的边
要么割掉用户带来的收益要么割掉成本

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N=60005,M=N*6,S=N-1,T=N-2,INF=1e9;int n=0,m=0;int head[N],to[M],c[M],f[M],next[M],edge=0;int ans=0;int q[M],front=0,back=0,d[N],cur[N];inline void addEdge(int u,int v,int ac) {    c[edge]=ac,to[edge]=v,next[edge]=head[u],head[u]=edge++;    c[edge]=0,to[edge]=u,next[edge]=head[v],head[v]=edge++;}inline void build() {    memset(head,-1,sizeof(head));    for (int i=1;i<=n;++i) {        int p=0;        scanf("%d",&p);        addEdge(m+i,T,p);    }    for (int i=1;i<=m;++i) {        int a=0,b=0,c2=0;        scanf("%d %d %d",&a,&b,&c2);        ans+=c2;        addEdge(S,i,c2);        addEdge(i,m+a,INF);        addEdge(i,m+b,INF);    }}inline int bfs() {    front=back=0;    memset(d,-1,sizeof(d));    d[S]=1;    q[back++]=S;    while (front<back) {        int x=q[front++];        for (int e=head[x];~e;e=next[e]) {            int v=to[e];            if (d[v]==-1 && c[e]>f[e]) {                d[v]=d[x]+1;                q[back++]=v;            }        }    }       return d[T]!=-1;}inline int dfs(int x,int t) {    if (x==T || !t)        return t;    int flow=0;    for (int e=cur[x];~e;e=next[e]) {        int v=to[e];        if (d[v]==d[x]+1 && c[e]>f[e]) {            int f0=dfs(v,min(t,c[e]-f[e]));            flow+=f0;            t-=f0;            f[e]+=f0;            f[e^1]-=f0;            if (c[e]>f[e])                cur[x]=e;            if (!t)                return flow;        }    }    if (t)        d[x]=-1;    return flow;}inline void dinic() {    while (bfs()) {        for (int e=1;e<=S;++e)            cur[e]=head[e];        ans-=dfs(S,INF);    }}int main(void) {    scanf("%d %d",&n,&m);    build();    dinic();    printf("%d\n",ans);         return 0;}

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