CSU 1506 Double Shortest Paths

1506: Double Shortest Paths

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 49  Solved: 5

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 4
1 2 5 1
2 4 6 0
1 3 4 0
3 4 9 1
4 4
1 2 5 10
2 4 6 10
1 3 4 10
3 4 9 10

Sample Output

Case 1: 23
Case 2: 24

HINT

 

Source

湖南省第十届大学生计算机程序设计竞赛

解题：费用流

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct arc{
     int to,flow,cost,next;
     arc(int x = ,int y = ,int z = ,int nxt = -){
         to = x;
         flow = y;
         cost = z;
         next = nxt;
     }
 };
 arc e[maxn*maxn];
 int head[maxn],d[maxn],p[maxn];
 int tot,S,T;
 void add(int u,int v,int flow,int cost){
     e[tot] = arc(v,flow,cost,head[u]);
     head[u] = tot++;
     e[tot] = arc(u,,-cost,head[v]);
     head[v] = tot++;
 }
 bool in[maxn];
 bool spfa(){
     queue<int>q;
     for(int i = S; i <= T; ++i){
         p[i] = -;
         in[i] = false;
         d[i] = INF;
     }
     d[S] = ;
     q.push(S);
     while(!q.empty()){
         int u = q.front();
         q.pop();
         in[u] = false;
         for(int i = head[u]; ~i; i = e[i].next){
             if(e[i].flow && d[e[i].to] > d[u] + e[i].cost){
                 d[e[i].to] = d[u] + e[i].cost;
                 p[e[i].to] = i;
                 if(!in[e[i].to]){
                     in[e[i].to] = true;
                     q.push(e[i].to);
                 }
             }
         }
     }
     return p[T] > -;
 }
 int solve(){
     int ans = ;
     while(spfa()){
         int minF = INF;
         for(int i = p[T]; ~i; i = p[e[i^].to])
             minF = min(minF,e[i].flow);
         for(int i = p[T]; ~i; i = p[e[i^].to]){
             e[i].flow -= minF;
             e[i^].flow += minF;
         }
         ans += d[T]*minF;
     }
     return ans;
 }
 int main(){
     int n,m,u,v,ai,di,cs = ;
     while(~scanf("%d %d",&n,&m)){
         memset(head,-,sizeof(head));
         S = tot = ;
         T = n + ;
         for(int i = ; i < m; ++i){
             scanf("%d %d %d %d",&u,&v,&ai,&di);
             add(u,v,,ai);
             add(u,v,,ai+di);
         }
         add(S,,,);
         add(n,T,,);
         printf("Case %d: %d\n",cs++,solve());
     }
     return ;
 }