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
解题：费用流
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 1000; 18 struct arc{ 19 int to,flow,cost,next; 20 arc(int x = 0,int y = 0,int z = 0,int nxt = -1){ 21 to = x; 22 flow = y; 23 cost = z; 24 next = nxt; 25 } 26 }; 27 arc e[maxn*maxn]; 28 int head[maxn],d[maxn],p[maxn]; 29 int tot,S,T; 30 void add(int u,int v,int flow,int cost){ 31 e[tot] = arc(v,flow,cost,head[u]); 32 head[u] = tot++; 33 e[tot] = arc(u,0,-cost,head[v]); 34 head[v] = tot++; 35 } 36 bool in[maxn]; 37 bool spfa(){ 38 queue<int>q; 39 for(int i = S; i <= T; ++i){ 40 p[i] = -1; 41 in[i] = false; 42 d[i] = INF; 43 } 44 d[S] = 0; 45 q.push(S); 46 while(!q.empty()){ 47 int u = q.front(); 48 q.pop(); 49 in[u] = false; 50 for(int i = head[u]; ~i; i = e[i].next){ 51 if(e[i].flow && d[e[i].to] > d[u] + e[i].cost){ 52 d[e[i].to] = d[u] + e[i].cost; 53 p[e[i].to] = i; 54 if(!in[e[i].to]){ 55 in[e[i].to] = true; 56 q.push(e[i].to); 57 } 58 } 59 } 60 } 61 return p[T] > -1; 62 } 63 int solve(){ 64 int ans = 0; 65 while(spfa()){ 66 int minF = INF; 67 for(int i = p[T]; ~i; i = p[e[i^1].to]) 68 minF = min(minF,e[i].flow); 69 for(int i = p[T]; ~i; i = p[e[i^1].to]){ 70 e[i].flow -= minF; 71 e[i^1].flow += minF; 72 } 73 ans += d[T]*minF; 74 } 75 return ans; 76 } 77 int main(){ 78 int n,m,u,v,ai,di,cs = 1; 79 while(~scanf("%d %d",&n,&m)){ 80 memset(head,-1,sizeof(head)); 81 S = tot = 0; 82 T = n + 1; 83 for(int i = 0; i < m; ++i){ 84 scanf("%d %d %d %d",&u,&v,&ai,&di); 85 add(u,v,1,ai); 86 add(u,v,1,ai+di); 87 } 88 add(S,1,2,0); 89 add(n,T,2,0); 90 printf("Case %d: %d ",cs++,solve()); 91 } 92 return 0; 93 }