UVA 11748 - Rigging Elections(dfs)
UVA 11748 - Rigging Elections
题意:n个人选举,给出m个人的投票人对于每一个人的优先级,如今你想让第c个人赢。问能不能
思路:对于两个人上场,假设a能赢b,就建一条a->b的边,然后问题事实上就变成是否能以c为根节点是一棵树,直接dfs一遍就可以
代码:
#include <cstdio> #include <cstring> #include <vector> using namespace std; const int N = 105; int n, m, c, a[N][N], vis[N]; vector<int> g[N]; int dfs(int u) { vis[u] = 1; int ans = 1; for (int i = 0; i < g[u].size(); i++) if (!vis[g[u][i]]) ans += dfs(g[u][i]); return ans; } int main() { while (~scanf("%d%d%d", &n, &m, &c) && n) { int u; for (int i = 1; i <= n; i++) g[i].clear(); for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++) { scanf("%d", &u); a[i][u] = j; } for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { int cnt = 0; for (int k = 1; k <= m; k++) { if (a[k][i] < a[k][j]) cnt++; if (cnt > m / 2) break; } if (cnt > m / 2) g[i].push_back(j); else g[j].push_back(i); } } memset(vis, 0, sizeof(vis)); printf("%s\n", dfs(c) == n ? "yes" : "no"); } return 0; }