題目鏈接

題意

有N個騎士，每個騎士有自己不喜歡的人，你需要選擇奇數個騎士開一個圓桌會議，每個騎士不能和自己不喜歡的人挨著坐。
問需要刪除那些人，這些人不能組成圓桌會議

思路

• 按照補圖建邊求雙連通分量，然后判斷雙連通分量中是否有奇圈
• 判斷是否存在奇圈可以判斷二分圖，如果不存在二分圖則存在奇圈
  統計可以組成圓桌會議的人，總數減去即可

#include <iostream> #include <queue> #include <stdio.h> #include <cstring> #include <vector> const int maxn = 1e3 + 5; const int inf = 0x3f3f3f3f; using namespace std; vector<int> g[maxn]; int gg[maxn][maxn]; int dfn[maxn], low[maxn], vis[maxn], Stack[maxn], inStack[maxn]; int color[maxn]; int len, cnt, ts; int n, m; void init(int n) {len = cnt = ts = 0;for (int i = 1; i <= n; ++i) g[i].clear();fill(vis, vis+n+1, 0);fill(dfn, dfn+n+1, 0);fill(gg[0], gg[0]+maxn*maxn, 0); } int check() {fill(color, color+maxn, -1);queue<int> que;for (int i = 1; i <= n; ++i) {if (inStack[i]) {que.push(i);color[i] = 0;break;}}while (!que.empty()) {int u = que.front();que.pop();for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (inStack[v] == 0) continue;if (color[v] == -1) {color[v] = color[u] ^ 1;que.push(v);}else if (color[v] == color[u]) return 1;}}return 0; } void tarjan(int u, int fa) {dfn[u] = low[u] = ++ts;Stack[len++] = u;for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i];if (v == fa) continue;if (!dfn[v]) {tarjan(v, u);low[u] = min(low[u], low[v]);if (dfn[u] <= low[v]) {fill(inStack, inStack+n+1, 0);inStack[u] = 1; while (1) {int top = Stack[--len];inStack[top] = 1;if (top == v) break;}if (check()) {for (int i = 1; i <= n; ++i) {if (inStack[i]) vis[i] = 1;}}}}else low[u] = min(low[u], dfn[v]);} } int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);while (scanf("%d %d", &n, &m), n) {init(n);for (int i = 0; i < m; ++i) {int u, v;scanf("%d%d", &u, &v);gg[u][v] = gg[v][u] = 1;} for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) {if (i == j || gg[i][j]) continue;g[i].push_back(j);}}for (int i = 1; i <= n; ++i) {if (dfn[i]) continue;tarjan(i, 0);}int ans = n;for (int i = 1; i <= n; ++i) ans -= vis[i];printf("%d\n", ans);}return 0; }

總結

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