标签:ret 需要 uva 技巧 main == tor stream names
点连通度:最少删除几个点使图不连通
拆点就变成了最小割
注意编号。画图就知道u’连v,v’连u。
技巧:不需要枚举S,T。固定S,枚举T即可
这种输入很烦, scanf(" (%d,%d)", &u, &v);
Scanf中添加
空白字符: 空白字符会使scanf()函数在读操作中略去输入中的一个或多个空白字符。
非空白字符: 一个非空白字符会使scanf()函数在读入时剔除掉与这个非空白字符相同的字符。
学到了
#include <cstdio> #include <queue> #include <cstring> #include <vector> #include <iostream> using namespace std; const int maxn = 100 + 5; const int INF = 0x3f3f3f3f; struct Edge { int from, to, cap, flow; Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {} }; struct EdmondsKarp { int n, m; vector<Edge> edges; vector<int> G[maxn]; int a[maxn]; int p[maxn]; void init(int n) { for (int i = 0; i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } int Maxflow(int s, int t) { int flow = 0; for (;;) { memset(a, 0, sizeof(a)); queue<int> Q; Q.push(s); a[s] = INF; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i < G[x].size(); i++) { Edge& e = edges[G[x][i]]; if (!a[e.to] && e.cap > e.flow) { p[e.to] = G[x][i]; a[e.to] = min(a[x], e.cap - e.flow); Q.push(e.to); } } if (a[t]) break; } if (!a[t]) break; for (int u = t; u != s; u = edges[p[u]].from) { edges[p[u]].flow += a[t]; edges[p[u] ^ 1].flow -= a[t]; } flow += a[t]; } return flow; } }; EdmondsKarp g; vector<Edge> bak; //恢复 int main() { int n,m; int u,v; while (scanf("%d%d", &n, &m) == 2) { g.init(n * 2); for (int i = 0; i < n; i++) g.AddEdge(i, i + n, 1); for (int i = 0; i < m; i++) { scanf(" (%d,%d)", &u, &v); //学到了 g.AddEdge(u + n, v, INF); g.AddEdge(v + n, u, INF); } int s = n, flow = n; bak = g.edges; for (int i = 1; i < n; i++) { //枚举终点 g.edges = bak; flow = min(flow, g.Maxflow(s, i)); } printf("%d\n", flow); } return 0; } /* 5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)*/
标签:ret 需要 uva 技巧 main == tor stream names
原文地址:https://www.cnblogs.com/lqerio/p/9860937.html