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 intusing 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0void 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0edge[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 3510 20 1010 10 3010 40 105 5 01 10 10001 10 10001 10 10000 10 10001 10 10000 0 0ans:30040*/