Luogu1264 K-联赛

这题其实不难想到

Description

link

题意太长了，概括不来，去题库里扫一眼吧（但是很好懂）

Solution

[Begin ]

考虑一个事情：每一个队伍的输局是没有用的

贪心一下，让每个队伍把剩下的比赛赢下来的时候，最有可能夺冠

设最终当前队赢得的场数的 (maxx)

接下来我们考虑让剩下的比赛怎么胜负

非这个队在剩下的比赛中的最大胜场数不得大于 (maxx-w_j) ，其中$j in[1,n] $且 (j!=i)

然后我们建图

每一场比赛都会让胜者的胜利场次加 (1) ,把 比赛的场次当成一个点，队伍当成一个点

每场比赛向队伍分别连边，边权为 (1) ，源点向 比赛场次连，然后队伍向汇点连

跑网络流就行了（或者这玩意就是个二分图完美匹配？？）

[Finish ]

Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define cl(x) memset(x, 0, sizeof(x))
namespace yspm {
inline int read() {
    int res = 0, f = 1;
    char k;
    while (!isdigit(k = getchar()))
        if (k == '-')
            f = -1;
    while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
    return res * f;
}
int n, s, t, tot;
const int N = 1e4 + 10;
int head[N], w[N], a[N][N], id[N][N], dep[N], cnt = 1;
struct node {
    int nxt, to, lim;
} e[N << 1];
inline void add2(int u, int v, int w) {
    e[++cnt].lim = w;
    e[cnt].nxt = head[u];
    e[cnt].to = v;
    return head[u] = cnt, void();
}
inline void add1(int u, int v, int w) {
    add2(u, v, w);
    add2(v, u, 0);
    return;
}
queue<int> q;
inline bool bfs() {
    cl(dep);
    dep[s] = 1;
    q.push(s);
    while (q.size()) {
        int fr = q.front();
        q.pop();
        for (int i = head[fr]; i; i = e[i].nxt) {
            int t = e[i].to;
            if (!dep[t] && e[i].lim)
                dep[t] = dep[fr] + 1, q.push(t);
        }
    }
    return dep[t];
}
inline int dfs(int now, int in) {
    if (now == t)
        return in;
    for (int i = head[now]; i && in; i = e[i].nxt) {
        int t = e[i].to;
        if (!e[i].lim || dep[t] != dep[now] + 1)
            continue;
        int res = dfs(t, min(e[i].lim, in));
        e[i].lim -= res;
        e[i ^ 1].lim += res;
        if (res)
            return res;
    }
    return 0;
}
inline void solve(int x) {
    cl(head);
    cnt = 1;
    cl(e);
    int maxx = w[x], tmp = 0;
    for (int i = 1; i <= n; ++i) maxx += a[x][i];
    for (int i = 1; i <= n; ++i) {
        if (i == x)
            continue;
        if (w[i] > maxx)
            return;
        add1(i, t, maxx - w[i]);
        for (int j = 1; j < i; ++j) {
            if (j != x && a[i][j]) {
                add1(s, id[i][j], a[i][j]), tmp += a[i][j];
                add1(id[i][j], i, a[i][j]);
                add1(id[i][j], j, a[i][j]);
            }
        }
    }
    int sum = 0, d;
    while (bfs()) {
        while (d = dfs(s, 1e15 + 10)) sum += d;
    }
    if (sum == tmp)
        printf("%lld ", x);
    return;
}
signed main() {
    n = read();
    s = n + 1, t = n + 2, tot = n + 2;
    for (int i = 1, k; i <= n; ++i) w[i] = read(), k = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j) a[i][j] = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j) id[i][j] = ++tot;
    for (int i = 1; i <= n; ++i) solve(i);
    puts("");
    return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }