HDU-3478Catch二分图的否命题

HDU-3478Catch

题意：

考虑Thief能否；

由于我推着推着就想到必须要三点可以互通，和二分图的结论正好相反，所以就试了一发，

真没想到thief的初始位置是不用考虑的。

下面是ac代码：

#include <bits/stdc++.h>
using namespace std;
int n, m, s, book[100000 + 10];
vector<int> mp[100000 + 10];
bool dfs(int u) {
    queue<int> q;
    q.push(u);
    book[u] = 1;
    while (!q.empty()) {
        int from = q.front();
        q.pop();
        for (int i = 0; i < mp[from].size(); i++) {
            int tmp = mp[from][i];
            if (book[tmp] == -1) {
                book[tmp] = !book[from];
                q.push(tmp);
            }

            else if (book[tmp] == book[from])
                return false;
        }
    }
    return true;
}
void init() {
    for (int i = 0; i < n; i++)
        mp[i].clear();
    memset(book, -1, sizeof(book));
}
int main() {
    int ca = 0, t;
    scanf("%d", &t);
    while (t--) {
        init();
        scanf("%d%d%d", &n, &m, &s);
        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            mp[u].push_back(v);
            mp[v].push_back(u);
        }
        bool flag = false;
        for (int i = 0; i < n && !flag; i++) {
            if (book[i] == -1 && !dfs(i)) {
                flag = true;
                break;
            }
        }

        if (flag)
            printf("Case %d: YES\n", ++ca);
        else
            printf("Case %d: NO\n", ++ca);
    }

    return 0;
}