题面

luogu

题解

最小费用最大流

先假设剩下\(m\)场比赛，双方全输。
考虑\(i\)赢一局的贡献
\(C_i*(a_i+1)^2+D_i*(b_i-1)^2-C_i*a_i^2-D_i*b_i^2\)
\(=C _i+2*a_i*C_i+D_i-2*b_i*D_i\)

建\(m\)个点限制每场比赛只有一个人赢，自\(S\)连一条\((1, 0)\)的边

然后从这\(m\)的点连向和比赛有关的两个点一条\((1, 0)\)的边

考虑关于\(t\)的边

因为\(a_i\)和\(b_i\)会变，不能直接连一条边

然后我们观察一下上面那个式子，
显然，赢得更多，贡献就增长越大

那么，拆边，我们可以对于每个点连向\(t\)很多条边，容量为\(1\)，价值是赢了对应场数的贡献

贪心地想，因为赢得更多，贡献增长越大，要最小费用，我们会先走赢得少的边
这样累加答案刚好满足条件呢

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}

const int N = 10000, inf = 2147483647;

struct node {
    int to, nxt, w, v;
}g[200*N];
int last[N], gl = 1;
void add(int x, int y, int w, int v) {
    g[++gl] = (node) {y, last[x], w, v};
    last[x] = gl;
    g[++gl] = (node) {x, last[y], 0, -v};
    last[y] = gl;
}

queue<int> q;
int pre[N], dis[N], from[N], s, t;
bool vis[N];

bool spfa() {
    memset(dis, 127, sizeof(dis));
    dis[s] = 0;
    q.push(s);
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = last[u]; i; i = g[i].nxt) {
            int v = g[i].to;
            if (g[i].w && dis[v] > dis[u] + g[i].v) {
                dis[v] = dis[u] + g[i].v; pre[v] = u; from[v] = i;
                if (!vis[v]) {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
        vis[u] = 0;
    }
    return dis[t] != dis[0];
}

int Mcmf() {
    int ans = 0;
    while (spfa()) {
        int di = inf;
        for (int i = t; i != s; i = pre[i]) di = min(di, g[from[i]].w);
        ans += di*dis[t];
        for (int i = t; i != s; i = pre[i]) g[from[i]].w -= di, g[from[i]^1].w += di;
    }
    return ans;
}
int a[N], b[N], c[N], d[N], cnt[N];

int main() {
    int n, m, sum = 0;
    read(n), read(m);
    s = m+n+1, t = s+1;
    for (int i = 1; i <= n; i++)
        read(a[i]), read(b[i]), read(c[i]), read(d[i]);
    for (int i = 1; i <= m; i++) {
        int x, y;
        read(x), read(y);
        b[x]++, b[y]++;
        cnt[x]++; cnt[y]++;
        add(s, i, 1, 0);
        add(i, x+m, 1, 0);
        add(i, y+m, 1, 0);
    }
    for (int i = 1; i <= n; i++)
        sum += c[i]*a[i]*a[i] + d[i]*b[i]*b[i];
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= cnt[i]; j++) {
            add(i+m, t, 1, c[i]+2*a[i]*c[i]+d[i]-2*b[i]*d[i]);
            b[i]--; a[i]++;
        }
    }
    printf("%d\n", sum+Mcmf());
    return 0;
}