标签：source lines clr com script direct for pre ble
Problem Description
Input
The first line of input contains an integer T, denoting the number of test cases.For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output
Case 1: 1 Case 2: 2
Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define CLR(x , v) memset(x , v , sizeof(x)) 5 6 static const int OO = 0x3fffffff; 7 8 static const int MAXN = 1e3 + 10; 9 10 int n , m; 11 int cap[MAXN][MAXN]; 12 int flow[MAXN][MAXN]; 13 int add[MAXN]; 14 queue<int> q; 15 int pre[MAXN]; 16 int main() 17 { 18 int t; 19 scanf("%d" , &t); 20 for(int c = 1 ; c <= t ; ++c) 21 { 22 int ans = 0; 23 CLR(cap , 0); 24 CLR(flow , 0); 25 CLR(pre , 0); 26 while(!q.empty()) 27 q.pop(); 28 scanf("%d%d" , &n , &m); 29 for(int i = 1 ; i <= m ; ++i) 30 { 31 int a , b , c; 32 scanf("%d%d%d" , &a , &b , &c); 33 cap[a][b] += c; 34 } 35 36 for(;;) 37 { 38 CLR(add , 0); 39 add[1] = OO; 40 q.push(1); 41 while(!q.empty()) 42 { 43 int u = q.front(); 44 q.pop(); 45 for(int i = 1 ; i <= n ; ++i) 46 { 47 if(!add[i] && cap[u][i] > flow[u][i]) 48 { 49 pre[i] = u; 50 q.push(i); 51 add[i] = min(add[u] , cap[u][i] - flow[u][i]); 52 } 53 } 54 } 55 56 if(!add[n]) 57 break; 58 59 for(int i = n ; i != 1 ; i = pre[i]) 60 { 61 flow[pre[i]][i] += add[n]; 62 flow[i][pre[i]] -= add[n];//反向弧 63 } 64 65 ans += add[n]; 66 } 67 68 printf("Case %d: %d\n" , c , ans); 69 } 70 }
标签：source lines clr com script direct for pre ble
原文地址：http://www.cnblogs.com/jianglingxin/p/6932641.html
