POJ 1966 Cable TV Network

无向图的顶点连通度求解。

注意：枚举顶点时，需要将网络流重新置0，不然会WA。

具体的求解方法请见我的另一篇博客：http://blog.csdn.net/wall_f/article/details/8210146

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <string>using namespace std;const int MAXN = 105;const int MAXM = 105*105;const int INF = 0x3f3f3f3f;struct Edge{int v, next;int f;}edge[MAXM];int cnt;int n, m;int first[MAXN], level[MAXN];int q[MAXN];void init(){cnt = 0;memset(first, -1, sizeof(first));}void read_graph(int u, int v, int f){edge[cnt].v = v, edge[cnt].f = f;edge[cnt].next = first[u], first[u] = cnt++;edge[cnt].v = u, edge[cnt].f = 0;edge[cnt].next = first[v], first[v] = cnt++;}int bfs(int s, int t){memset(level, 0, sizeof(level));level[s] = 1;int front = 0, rear = 1;q[front] = s;while(front < rear){int x = q[front++];if(x == t) return 1;for(int e = first[x]; e != -1; e = edge[e].next){int v = edge[e].v, f = edge[e].f;if(!level[v] && f){level[v] = level[x] + 1;q[rear++] = v;}}}return 0;}int dfs(int u, int maxf, int t){if(u == t) return maxf;int ret = 0;for(int e = first[u]; e != -1; e = edge[e].next){int v = edge[e].v, f = edge[e].f;if(level[v] == level[u] + 1 && f){int Min = min(maxf-ret, f);f = dfs(v, Min, t);edge[e].f -= f;edge[e^1].f += f;ret += f;if(ret == maxf) return ret;}}return ret;}int Dinic(int s, int t){int ans = 0;while(bfs(s, t)) ans += dfs(s, INF, t);return ans;}void init_flow() //重置网络流为初始值{for(int i = 0; i < cnt; i += 2){edge[i].f += edge[i^1].f;edge[i^1].f = 0;}}void read_case(){init();for(int i = 0; i < n; i++){read_graph(i, i+n, 1);}for(int i = 0; i < m; i++){int u, v;while(getchar() != '(' ) ;scanf("%d,%d)", &u, &v);read_graph(u+n, v, INF);read_graph(v+n, u, INF);}}void solve(){read_case();int ans = INF;for(int i = 1; i < n; i++){ans = min(ans, Dinic(0+n, i)); //以0''为源点，枚举其他顶点i' init_flow();}if(ans == INF) ans = n;printf("%d\n", ans);}int main(){while(~scanf("%d%d", &n, &m)){solve();}return 0;}