CSU 1506: Double Shortest Paths（最小费用流）（湖南省第十届省赛）

1506: Double Shortest Paths

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 84  Solved: 22
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Description

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

Source

湖南省第十届大学生计算机程序设计竞赛

两个人从起点走到终点  第一个人走的时候所花分费用为边上第一个值  第二个人走的时候所花的费用为第一个值加上第二个值

求两个人走到终点所花的最小费用

对于u->v 建两条边 容量都为1 第一个的费用为第一个值 第二个的费用为两个值的和

另外再加上超级源点到起点的边 终点到超级汇点的边

最小费用流求出最小费用就可以了

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 100000#define MAXM 100000#define INF 0x3f3f3f3f#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;void read(int &x){    char ch;    x=0;    while(ch=getchar(),ch!=' '&&ch!='\n')    {        x=x*10+ch-'0';    }}struct Edge{    int to,next,cap,flow,cost;}edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N;void init(int n){    N=n;    tol=0;    MEM(head,-1);}void addedge(int u,int v,int cap,int cost){    edge[tol].to=v;    edge[tol].cap=cap;    edge[tol].cost=cost;    edge[tol].flow=0;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].to=u;    edge[tol].cap=0;    edge[tol].cost=-cost;    edge[tol].flow=0;    edge[tol].next=head[v];    head[v]=tol++;}bool spfa(int s,int t){    queue<int> q;    for(int i=0;i<N;i++)    {        dis[i]=INF;        vis[i]=false;        pre[i]=-1;    }    dis[s]=0;    vis[s]=1;    q.push(s);    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=false;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)            {                dis[v]=dis[u]+edge[i].cost;                pre[v]=i;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }    if(pre[t]==-1)  return false;    return true;}int minCostMaxflow(int s,int t,int &cost){    int flow=0;    cost=0;    while(spfa(s,t))    {        int Min=INF;        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])        {            if(Min>edge[i].cap-edge[i].flow)                Min=edge[i].cap-edge[i].flow;        }        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])        {            edge[i].flow+=Min;            edge[i^1].flow-=Min;            cost+=edge[i].cost*Min;        }        flow+=Min;    }    return flow;}int main(){//    fread;    int n,m;    int cs=1;    while(scanf("%d%d",&n,&m)!=EOF)    {        init(n+2);        for(int i=0;i<m;i++)        {            int u,v,c1,c2;            scanf("%d%d%d%d",&u,&v,&c1,&c2);            addedge(u,v,1,c1);            addedge(u,v,1,c1+c2);        }        int s=0,t=n+1;        addedge(0,1,2,0);        addedge(n,t,2,0);        int cost;        minCostMaxflow(s,t,cost);        printf("Case %d: %d\n",cs++,cost);    }    return 0;}