HDU3861-The King’s Problem(有向圖強連通縮點+最小路徑覆蓋)
阿新 • • 發佈:2019-01-23
題意:題目大意:一個有向圖,讓你按規則劃分區域,要求劃分的區域數最少。
規則如下:
1、有邊u到v以及有邊v到u,則u,v必須劃分到同一個區域內。
2、一個區域內的兩點至少要有一方能到達另一方。
3、一個點只能劃分到一個區域內。
思路:根據規則1可知必然要對強連通分量進行縮點,縮點後變成了一個弱連通圖。根據規則2、3可知即是要求圖的最小路徑覆蓋。
程式碼:
#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; }