[代码]SGU 270 Thimbles

Abstract

SGU 270 Thimbles

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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;

}