首页 > 代码库 > hdu3861他的子问题是poj2762二分匹配+Tarjan+有向图拆点 其实就是求DAG的最小覆盖点
hdu3861他的子问题是poj2762二分匹配+Tarjan+有向图拆点 其实就是求DAG的最小覆盖点
The King’s Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3120 Accepted Submission(s): 1096
Problem Description
In the Kingdom of Silence, the king has a new problem. There are N cities in the kingdom and there are M directional roads between the cities. That means that if there is a road from u to v, you can only go from city u to city v, but can’t go from city v to city u. In order to rule his kingdom more effectively, the king want to divide his kingdom into several states, and each city must belong to exactly one state. What’s more, for each pair of city (u, v), if there is one way to go from u to v and go from v to u, (u, v) have to belong to a same state. And the king must insure that in each state we can ether go from u to v or go from v to u between every pair of cities (u, v) without passing any city which belongs to other state.
Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.
Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.
Input
The first line contains a single integer T, the number of test cases. And then followed T cases.
The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.
The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.
Output
The output should contain T lines. For each test case you should just output an integer which is the least number of states the king have to divide into.
Sample Input
1
3 2
1 2
1 3
Sample Output
2
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; const int MAXN = 20010; const int MAXM = 100010; struct Edge{ int to, next; }edge[MAXM]; int head[MAXN], tot; int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN]; int Index, top; int scc; bool Instack[MAXN]; int num[MAXN]; int n, m; void init() { tot = 0; memset(head, -1, sizeof(head)); } void addedge(int u, int v) { edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } void Tarjan(int u) { int v; Low[u] = DFN[u] = ++Index; Stack[top++] = u; Instack[u] = true; for (int i = head[u]; i != -1; i = edge[i].next) { v = edge[i].to; if (!DFN[v]) { Tarjan(v); if (Low[u] > Low[v]) Low[u] = Low[v]; } else if (Instack[v] && Low[u] > DFN[v]) Low[u] = DFN[v]; } if (Low[u] == DFN[u]) { scc++; do { v = Stack[--top]; Instack[v] = false; Belong[v] = scc; num[scc]++; } while (v != u); } } void solve() { memset(Low, 0, sizeof(Low)); memset(DFN, 0, sizeof(DFN)); memset(num, 0, sizeof(num)); memset(Stack, 0, sizeof(Stack)); memset(Instack, false, sizeof(Instack)); Index = scc = top = 0; for (int i = 1; i <= n; i++) if (!DFN[i]) Tarjan(i); } vector<int> g[MAXN]; int linker[MAXN], used[MAXN]; bool dfs(int u) { for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (!used[v]) { used[v] = 1; if (linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return true; } } } return false; } int hungary() { int res = 0; memset(linker, -1, sizeof(linker)); for (int i = 1; i <= scc; i++) { memset(used, 0, sizeof(used)); if (dfs(i)) res++; } return scc - res; } int main() { int cas; scanf("%d", &cas); while (cas--) { scanf("%d%d", &n, &m); init(); int u, v; for (int i = 0; i < m; i++) { scanf("%d%d", &u, &v); addedge(u, v); } solve(); for (int i = 0; i <= scc; i++) g[i].clear(); for (int u = 1; u <= n; u++) { for (int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if (Belong[u] != Belong[v]) g[Belong[u]].push_back(Belong[v]); } } printf("%d\n", hungary()); } return 0; }
hdu3861他的子问题是poj2762二分匹配+Tarjan+有向图拆点 其实就是求DAG的最小覆盖点
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