uva 12821 Double Shortest Paths

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Description

最小费用最大流

Input

There will be at most 200 test cases. Each case begins with two integers n, m (1<=n<=500, 1<=m<=2000), the number of caves and passages. Each of the following m lines contains four integers u, v, di and ai (1<=u,v<=n, 1<=di<=1000, 0<=ai<=1000). Note that there can be multiple passages connecting the same pair of caves, and even passages connecting a cave and itself.

Output

For each test case, print the case number and the minimal total difficulty.

Sample Input

4 41 2 5 12 4 6 01 3 4 03 4 9 14 41 2 5 102 4 6 101 3 4 103 4 9 10

Sample Output

Case 1: 23Case 2: 24

HINT

这本是湖南省一次省赛中的题目，因为自己学校的oj无法提交了，只好在vua上提交了。。。

思路：在没学网络流之前还以为是个最短路问题，然后wa了n发，后来才知道是一个最小费用最大流问题。这里讲将困难度看做费用，再为每一条边上为1的流量，然后将1节点连到源点，流量为2，费用为0，将n连到汇点，流量为2，费用为0，这样求出最小费用最大流就是题目所求

#include <iostream>#include <stdio.h>#include <cstring>#include <vector>#include <queue>#define INF 0x3f3f3f3fusing namespace std;const int maxn = 666;struct Edge{int from,to,cap,flow,cost;Edge(){}Edge(int a,int b,int c,int d,int e):from(a),to(b),cap(c),flow(d),cost(e){}};struct MCMF{int n,m,s,t;vector<Edge> edges;vector<int> g[maxn];int inq[maxn];int d[maxn];int p[maxn];int a[maxn];void init(int n){this->n =n;for(int i=0;i<n;i++)g[i].clear();edges.clear();}void addedge(int from,int to,int cap,int cost){Edge e1= Edge(from,to,cap,0,cost), e2= Edge(to,from,0,0,-cost);edges.push_back(e1);edges.push_back(e2);m=edges.size();g[from].push_back(m-2);g[to].push_back(m-1);}bool spfa(int s,int t, int & flow,int & cost){for(int i=0;i<n;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=0;i<g[u].size();i++){Edge & e = edges[g[u][i]];if(e.cap>e.flow && d[e.to]>d[u]+e.cost){d[e.to]=d[u]+e.cost;p[e.to]=g[u][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+=a[t]*d[t];int u=t;while(u!=s){edges[p[u]].flow+=a[t];edges[p[u]^1].flow-=a[t];u=edges[p[u]].from;}return true;}int  Mincost(int s,int t){int flow=0,cost =0;while(spfa(s,t,flow,cost));return cost;}}sol;int main(){//freopen("in.txt","r",stdin);int n,m,cas=1;while(scanf("%d%d",&n,&m)!=EOF){    //printf("%d %d",n,m);int s=0,t=n+1;sol.init(n+2);while(m--){int u,v,c,ad;scanf("%d%d%d%d",&u,&v,&c,&ad);//printf(" %d%d%d%d\n",u,v,c,ad);sol.addedge(u,v,1,c);sol.addedge(u,v,1,c+ad);}sol.addedge(s,1,2,0);sol.addedge(n,t,2,0);printf("Case %d: %d\n",cas++,sol.Mincost(s,t));}}

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