[最大权闭合图]hdoj 3879：Base Station

大致题意：

有n个可以被修建的通信基站，修建每个基站都需要一定的花费。给出m组关系，每组关系由三个数a,b,c表示，代表在a和b之间修建通信路的话可以带来c的收益。求最大纯收益是多少。

大致思路：

把问题转化为最大全闭合图模型。原图中的点转化为闭合图中的负权点，原图中的边转化为闭合图中的正权点。剩下就是套最大权闭合图模版的问题了。

详细代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int nMax=60005;
const int mMax=500000;
const int inf=1<<30;
class node{
public:
int u,v,next;
int c;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax],ps[nMax],dep[nMax],n,m,ans;
void addedge(int u, int v,int c){    //
edge[ne].u = u;
edge[ne].v = v;
edge[ne].c = c;
edge[ne].next = head[u];
head[u] = ne ++;
edge[ne].u = v;
edge[ne].v = u;
edge[ne].c =0;
edge[ne].next = head[v];
head[v] = ne ++;
}

int dinic(int s, int t){     // dinic模板：源点为s，汇点为t
int tr, res = 0;
int i, j, k, f, r, top;
while(1){
memset(dep, -1, sizeof(dep));
for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
if(edge[j].c && dep[k=edge[j].v] == -1){
dep[k] = dep[i] + 1;
ps[r ++] = k;
if(k == t){
f = r; break;
}
}
if(dep[t] == -1) break;
memcpy(cur, head, sizeof(cur));
i = s, top = 0;
while(1){
if(i == t){
for(tr = inf, k = 0; k < top; k ++)
if(edge[ps[k]].c < tr)
tr = edge[ps[f=k]].c;
for(k = 0; k < top; k ++){
edge[ps[k]].c -= tr;
edge[ps[k]^1].c += tr;
}
i = edge[ps[top=f]].u;
res += tr;          //
}
for(j = cur[i]; cur[i]; j = cur[i] = edge[cur[i]].next)
if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
if(cur[i]){
ps[top ++] = cur[i];
i = edge[cur[i]].v;   //
}else{
if(top == 0) break;
dep[i] = -1;
i = edge[ps[-- top]].u;
}
}
}
return res;
}

int main(){
int ans,c,t,i,a,b,s;
while(scanf("%d%d",&n,&m)!=EOF){
ne=2;
ans=0;
s=0,t=n+m+1;
memset(head,0,sizeof(head));
for(i=1;i<=n;i++){
scanf("%d",&a);
addedge(s,i,a);
}
for(i=n+1;i<=n+m;i++){
scanf("%d%d%d",&a,&b,&c);
ans+=c;
addedge(a,i,inf);
addedge(b,i,inf);
addedge(i,t,c);
}
ans-=dinic(s,t);
printf("%d\n",ans);
}
return 0;
}