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Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 8218 Accepted Submission(s): 3824
Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
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
Author HyperHexagon
Source HyperHexagon's Summer Gift (Original tasks)
题意：
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纯求网络最大流，用EK算法就可以实现。
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代码：
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#include
#include
#include
#include
using namespace std; #define maxn 20 #define INF 0x3f3f3f3f int ans, s, t, n; int a[maxn], pre[maxn]; int flow[maxn][maxn]; int cap[maxn][maxn]; void Edmonds_Karp() { queue
q; memset(flow, 0, sizeof(flow)); ans = 0; while(1) { memset(a, 0, sizeof(a)); a[s] = INF; q.push(s); while(!q.empty()) //bfs找增光路径 { int u = q.front(); q.pop(); for(int v = 1; v <= n; v++) if(!a[v] && cap[u][v] > flow[u][v]) { pre[v] = u; q.push(v); a[v] = min(a[u], cap[u][v]-flow[u][v]); } } if(a[t] == 0) break; for(int u = t; u != s; u = pre[u]) //改进网络流 { flow[pre[u]][u] += a[t]; flow[u][pre[u]] -= a[t]; } ans += a[t]; } } int main() { //freopen(hdu_3549.txt, r, stdin); int T, m, u, v, c; scanf(%d, &T); for(int cas = 1; cas <= T; cas++) { scanf(%d%d, &n, &m); memset(cap, 0, sizeof(cap)); while(m--) { scanf(%d%d%d, &u, &v, &c); cap[u][v] += c; } s = 1, t = n; Edmonds_Karp(); printf(Case %d: %d , cas, ans); } return 0; }
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