CFGYM 2013-2014 CT S01E03 D题 费用流模版题

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/12032453

题意：

n行， a房间的气球，b房间的气球

i行需要的气球，与a房的距离，b房的距离

求最小距离

 

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <algorithm>
#include <stdlib.h>

#define N 2000
#define M 10100
#define inf 107374182
#define ll int

using namespace std;
inline ll Min(ll a,ll b){return a>b?b:a;}
inline ll Max(ll a,ll b){return a>b?a:b;}

int n;
//双向边，注意RE 注意这个模版是 相同起末点的边 合并而不是去重
struct Edge{
	int from, to, flow, cap, nex, cost;
}edge[M*2];

int head[N], edgenum;//2个要初始化-1和0
void addedge(int u,int v,int cap,int cost){//网络流要加反向弧
	Edge E={u, v, 0, cap, head[u], cost};
	edge[edgenum]=E;
	head[u]=edgenum++;

	Edge E2={v, u, 0, 0, head[v], -cost}; //这里的cap若是单向边要为0
	edge[edgenum]=E2;
	head[v]=edgenum++;
}
int D[N], P[N], A[N];
bool inq[N];
bool BellmanFord(int s, int t, int &flow, int &cost){
	for(int i=0;i<=n+4;i++) D[i]= inf;

	memset(inq, 0, sizeof(inq));
	D[s]=0;  inq[s]=1; P[s]=0; A[s]=inf;

	queue<int> Q;
	Q.push( s );
	while( !Q.empty()){
		int u = Q.front(); Q.pop();
		inq[u]=0;
		for(int i=head[u]; i!=-1; i=edge[i].nex){
			Edge &E = edge[i];
			if(E.cap > E.flow && D[E.to] > D[u] +E.cost){
				D[E.to] = D[u] + E.cost ;
				P[E.to] = i;
				A[E.to] = Min(A[u], E.cap - E.flow);
				if(!inq[E.to]) Q.push(E.to) , inq[E.to] = 1;
			}
		}
	}
	if(D[t] == inf) return false;
	flow += A[t];
	cost += D[t] * A[t];
	int u = t;
	while(u != s){
		edge[P[u]].flow += A[t];
		edge[P[u]^1].flow -= A[t];
		u = edge[P[u]].from;
	}
	return true;
}

int Mincost(int s,int t){
	int flow = 0, cost = 0;
	while(BellmanFord(0, n+3, flow, cost));
	return cost;
}
int main(){
	int i,disa,disb,flow,maxa,maxb;
	while( scanf("%d %d %d",&n, &maxa, &maxb), n+maxb+maxa){
		memset(head,-1,sizeof(head));	edgenum=0;

		addedge(n+1, n+3, maxa, 0);
		addedge(n+2, n+3, maxb, 0);

		for(i=1;i<=n;i++){
			scanf("%d %d %d",&flow,&disa,&disb);
			addedge(0, i, flow, 0);
			addedge(i, n+1, flow, disa);
			addedge(i, n+2, flow, disb);
		}

		printf("%d\n",Mincost(0,n+3) );

	}
	return 0;
}
/*
3 15 35
10 20 10
10 10 30
10 40 10

5 5 0
1 10 1000
1 10 1000
1 10 1000
0 10 1000
1 10 1000


0 0 0

ans:
300
40

*/