[代码]SGU 270 Thimbles

Abstract

SGU 270 Thimbles

图论 技巧 打补丁

 

Body

非常tricky的problem，有幸1y。

简单来说就是考虑每个点和点1直接联通，间接联通，在不在环里的情况。然后点1自己还要特判。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int N, M;
bool vis[111];
int g[111][111];

bool dfs(int s, int t)
{
    vis[s] = 1;
    if (s == t) return 1;
    for (int u = 0; u < N; ++u)
    {
        if (!g[s][u] || vis[u]) continue;
        if (dfs(u, t)) return 1;
    }
    return 0;
}

bool cyc(int u)
{
    bool res;
    for (int v = 0; v < N; ++v)
    {
        if (!g[u][v]) continue;
        g[u][v]--; g[v][u]--;
        memset(vis, 0, sizeof(vis));
        res = dfs(u, v);
        g[u][v]++; g[v][u]++;
        if (res) return 1;
    }
    return 0;
}

bool hmr(int u)
{
    memset(vis, 0, sizeof(vis));
    if (!dfs(0, u)) return 0;
    int x = g[0][u];
    g[0][u] = g[u][0] = 0;
    memset(vis, 0, sizeof(vis));
    if (x&1 || dfs(0, u))
    {
        g[0][u] = g[u][0] = x;
        return 1;
    }
    bool res = cyc(0)||cyc(u);
    g[0][u] = g[u][0] = x;
    return res;
}


int main()
{
    int i, j, k, u, v;
    cin >> N >> M;
    while (M--)
    {
        cin >> u >> v;
        u--; v--;
        g[u][v]++; g[v][u]++;
    }
    bool mdk = 1;
    for (u = 1; u<N && mdk; ++u)
        if (g[0][u]) mdk = 0;
    int syk = 0;
    for (u = 1; u<N && !mdk; ++u)
    {
        if (!g[0][u]) continue;
        int x = g[0][u];
        if ((x&1)==0)
        {
            mdk = 1;
            break;
        }
        g[0][u] = g[u][0] = 0;
        memset(vis, 0, sizeof(vis));
        if (dfs(0, u))
        {
            g[0][u] = g[u][0] = x;
            mdk = 1;
            break;
        }
        if (x>1 && cyc(u)) mdk = 1;
        g[0][u] = g[u][0] = x;
        if (x>1) syk++;
    }
    mdk |= (syk>=2);
    if (mdk) cout<<"1 ";
    for (u = 1; u < N; ++u)
        if (hmr(u)) printf("%d ", u+1);
    cout << endl;
    return 0;
}