标签:
挺裸的最小费用最大流。。。
#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define op() clr(head,0);pt=edges;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-‘0‘,c=getchar();
return x;
}
const int nmax=405;
const int maxn=100005;
const int inf=0x7f7f7f7f;
struct edge{
int to,cap,cost;edge *next,*rev;
};
struct oo{
int cap,cost;
oo(int cap,int cost):cap(cap),cost(cost){}
};
edge edges[maxn<<1],*pt,*head[nmax],*p[nmax];
int d[nmax],a[nmax];bool inq[nmax];
void add(int u,int v,int d,int w){
pt->to=v;pt->cap=d;pt->cost=w;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d,int w){
add(u,v,d,w);add(v,u,0,-w);head[u]->rev=head[v];head[v]->rev=head[u];
}
oo mincost(int s,int t){
int flow=0,cost=0;
while(1){
clr(d,0x7f),d[s]=0;clr(inq,0);inq[s]=1;a[s]=inf;
queue<int>q;q.push(s);
while(!q.empty()){
int x=q.front();q.pop();inq[x]=0;
qwq(x) if(o->cap>0&&d[o->to]>d[x]+o->cost){
int to=o->to;d[to]=d[x]+o->cost;
a[to]=min(a[x],o->cap);p[to]=o;
if(!inq[to]) q.push(to),inq[to]=1;
}
}
if(d[t]==inf) break;
flow+=a[t];cost+=d[t]*a[t];
int x=t;
while(x!=s) p[x]->cap-=a[t],p[x]->rev->cap+=a[t],x=p[x]->rev->to;
}
return oo(flow,cost);
}
int main(){
op();
int n=read(),m=read(),u,v,d;
REP(i,2,n-1) adde(i,n+i,1,0);
rep(i,m) u=read(),v=read(),d=read(),u==1?adde(u,v,1,d):adde(u+n,v,1,d);
/*rep(i,n) {
qwq(i) printf("%d ",o->to);printf("\n");
}*/
oo o=mincost(1,n);
printf("%d %d\n",o.cap,o.cost);
return 0;
}
对于30%的数据,N ≤ 20,M ≤ 120。
对于100%的数据,N ≤ 200,M ≤ 20000。
标签:
原文地址:http://www.cnblogs.com/fighting-to-the-end/p/5662650.html