加减法

Posted on Edited on In

题目大意：在一个无向图里面每次可以选择一条边，使得这条边的一个端点加1，另一个端点减1

求\(\sum i\times val[i]\)

题目分析：

1
考虑在同一个联通块里面，可以随意分配名额，所以我们尽量使得编号大的权值最大就好

代码如下：

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

const int N = 2e5 + 66;

int to[N << 1], nex[N << 1], head[N], cnt;

inline void add_edge(int x, int y)
{
    to[++ cnt] = y;
    nex[cnt] = head[x];
    return (void)(head[x] = cnt);
}

vector <int> v[N];
int n, m, L, tot;
int vis[N], val[N], w[N];

inline void dfs(int x, int num)
{
    int i, y;
    w[num] += val[x], v[num].push_back(x);
    vis[x] = 1;
    for (i = head[x]; i; i = nex[i])
    {
        y = to[i];
        if (vis[y]) continue;
        dfs(y, num);
    }
    return;
}

inline int fuck(int x) 
{
    sort(v[x].begin(), v[x].end());
    int s = v[x].size(), ret(0), i, y;
    for (i = 0; i < s; ++ i)
    {
        y = v[x][i];
        val[y] = 1;
    }
    w[x] -= s;
    for (i = s - 1; i >= 0; -- i)
    {
        y = v[x][i];
        val[y] += min(L - 1, w[x]);
        if (val[y] != L) w[x] = 0;
        else w[x] -= (L - 1);
        if (w[x] <= 0) break;
    }
    
    for (i = 0; i < s; ++ i)
    {
        y = v[x][i];
        ret = ret + y * val[y];
    }
    return ret;
}

signed main()
{
    int i, j, x, y, res(0);
    
    n = read(), m = read(), L = read();
    for (i = 1; i <= n; ++ i) val[i] = read();
    
    for (i = 1; i <= m; ++ i) 
    {
        x = read(), y = read();
        add_edge(x, y), add_edge(y, x);
    }
    
    for (i = 1; i <= n; ++ i) if (! vis[i]) ++ tot, dfs(i, tot);
    for (i = 1; i <= tot; ++ i) res += fuck(i);
    put(res);
    return 0;
}