[ICPC2018西安L] Eventual … Journey

[ICPC2018西安L] Eventual … Journey - 结论

Description

N个车站，M条边，每个车站分为黑白两种，如果两个点颜色一样，可以花1的花费到达，或者两个点有边直接连接，问每个点可以到达的点的总花费的最小值

Solution

直接相连或者同类代价为 1，两点度数都为 0 代价为 3，其它情况代价为 2

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e5 + 5;

int n, m, a[N], c[2], d[2];
vector<int> g[N];

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        if (a[u] == a[v])
            continue;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    for (int i = 1; i <= n; i++)
        c[a[i]]++;

    for (int i = 1; i <= n; i++)
        if (g[i].size() == 0)
            d[a[i]]++;
    if (m == 0)
    {
        for (int i = 1; i <= n; i++)
            cout << n - 1 << " ";
        return 0;
    }

    for (int i = 1; i <= n; i++)
    {
        if (g[i].size() > 0)
        {
            cout << n - 1 + c[1 - a[i]] - g[i].size() << " ";
        }
        else
        {
            cout << n - 1 + c[1 - a[i]] + d[1 - a[i]] << " ";
        }
    }
}