CFGYM 2013-2014 CT S01E03 D题 费用流模版题

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



*/