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POJ2914 Minimum Cut 【全局最小割】(Stoer_Wagner)
Time Limit: 10000MS | Memory Limit: 65536K | |
Total Submissions: 7610 | Accepted: 3203 | |
Case Time Limit: 5000MS |
Description
Given an undirected graph, in which two vertices can be connected by multiple edges, what is the size of the minimum cut of the graph? i.e. how many edges must be removed at least to disconnect the graph into two subgraphs?
Input
Input contains multiple test cases. Each test case starts with two integers N and M (2 ≤ N ≤ 500, 0 ≤ M ≤ N × (N ? 1) ? 2) in one line, where N is the number of vertices. Following are M lines, each line contains M integersA, B and C (0 ≤ A, B < N, A ≠ B, C > 0), meaning that there C edges connecting vertices A and B.
Output
There is only one line for each test case, which contains the size of the minimum cut of the graph. If the graph is disconnected, print 0.
Sample Input
3 3 0 1 1 1 2 1 2 0 1 4 3 0 1 1 1 2 1 2 3 1 8 14 0 1 1 0 2 1 0 3 1 1 2 1 1 3 1 2 3 1 4 5 1 4 6 1 4 7 1 5 6 1 5 7 1 6 7 1 4 0 1 7 3 1
Sample Output
2 1 2
Source
Wang, Ying (Originator)
Chen, Shixi (Test cases)
题意:求裸的最小割。
题解:过程有点像求“最大生成树”,每次找到将最后的点(即割点值最小的点)隔离出去的最小割大小nowcut,然后用nowcut更新结果mincut,然后将最后两个点合并,方式是将最后一个点合并到倒数第二个点,边也都合并过去,就这样循环下去直到只剩下一个顶点,此时mincut即为所求。
#include <stdio.h> #include <string.h> #define inf 0x7fffffff #define maxn 502 int map[maxn][maxn], W[maxn], hash[maxn]; bool vis[maxn]; void getMap(int n, int m) { int i, u, v, c; memset(map, 0, sizeof(map)); for(i = 0; i < m; ++i){ scanf("%d%d%d", &u, &v, &c); map[u][v] += c; map[v][u] += c; } } int Stoer_Wagner(int n) { int minCut = inf, nowCut, now, pre, i, j; for(i = 0; i < n; ++i) hash[i] = i; while(n > 1){ nowCut = -1; now = 1; vis[hash[0]] = 1; pre = 0; for(i = 1; i < n; ++i){ W[hash[i]] = map[hash[0]][hash[i]]; vis[hash[i]] = 0; if(W[hash[i]] > nowCut){ nowCut = W[hash[i]]; now = i; } } for(j = 1; j < n; ++j){ vis[hash[now]] = 1; if(j == n - 1){ if(nowCut < minCut) minCut = nowCut; for(i = 0; i < n; ++i){ map[hash[pre]][hash[i]] += map[hash[now]][hash[i]]; map[hash[i]][hash[pre]] += map[hash[now]][hash[i]]; } hash[now] = hash[--n]; break; } pre = now; nowCut = -1; for(i = 1; i < n; ++i){ if(!vis[hash[i]]){ W[hash[i]] += map[hash[pre]][hash[i]]; if(W[hash[i]] > nowCut){ nowCut = W[hash[i]]; now = i; } } } } } return minCut; } void solve(int n) { printf("%d\n", Stoer_Wagner(n)); } int main() { freopen("stdin.txt", "r", stdin); int n, m; while(scanf("%d%d", &n, &m) == 2){ getMap(n, m); solve(n); } return 0; }
POJ2914 Minimum Cut 【全局最小割】(Stoer_Wagner)