反向贪心决定博弈+博弈论中条件改变人的决策满足单调性：1229A

本文搬运自本人高中时期CSDN博客，若图片加载不出来，可到原文查看：https://blog.csdn.net/zhangtingxiqwq/article/details/135329131

http://cplusoj.com/d/senior/p/SS231229A

结论：

人们倒着来，每个人去掉当前对自己最不利的

因此我们有了 O(n3)O(n^3)

考虑当一个人变了后，每个人的决策必然满足单调性，因此就可以平方了

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start coding at 20:19
passing at 21:18
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// ubsan: undefined
// accoders
#include <bits/stdc++.h>
#include <random>
using namespace std;
#ifdef LOCAL
#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
#define debug(...) void(0)
#endif
//#define int long long
inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
#define Z(x) (x) * (x)
#define pb push_back
#define fi first
#define se second
// srand(time(0));
#define N 5010
//#define M
//#define mo
int a[N][N];
void gen(int n, int seed) {
    if (seed == 0) {
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n + 1; ++j) a[i][j] = read();
    } else {
        std::mt19937 rnd(seed);
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n + 1; ++j) {
                a[i][j] = j;
                std::swap(a[i][j], a[i][rnd() % j + 1]);
            }
        }
    }
}
struct node {
	int id, x; 
	bool operator < (const node &A) const {
		return x < A.x; 
	}
};
set<node>s[N]; 
int n, m, i, j, k, T, seed;
int ans[N], l, lst, b[N], cho[N]; 

int nxt(int x) {
	if(x == 1) return n ; 
	return x - 1; 
}

signed main() {
    freopen("sing.in", "r", stdin);
    freopen("sing.out", "w", stdout);
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    //	T=read();
    //	while(T--) {
    //
    //	}
    n = read();
    seed = read();
    gen(n, seed);
  	for(i=n; i>=1; --i) {
  		for(j=1; j<=n+1; ++j) if(!b[j] && (!cho[i] || a[i][j] < a[i][cho[i]])) cho[i]=j; 
  		b[cho[i]]=1; 
//  		debug("%d : %d\n", i, cho[i]); 
	}
//	for(j=1; j<=n; ++j) debug("%d ", cho[j]); debug("\n"); 
	for(j=1; j<=n+1; ++j) if(!b[j]) break; 
	printf("%d ", j); 
	for(i=2, l=n; i<=n; ++i, --l) {
		b[cho[l]] = 0; 
		for(j=nxt(l), lst = cho[l]; j != l; j = nxt(j)) 
			if(a[j][lst] < a[j][cho[j]]) {
				b[lst] = 1; b[cho[j]] = 0; 
				swap(lst, cho[j]); 
			}
		cho[l]=0; 
		for(j=1; j<=n+1; ++j) if(!b[j] && (!cho[l] || a[l][j] < a[l][cho[l]])) cho[l]=j; 
		b[cho[l]]=1; 
		for(j=1; j<=n+1; ++j) if(!b[j]) break; 
		ans[l]=j; 
	}
	for(i=2; i<=n; ++i) printf("%d ", ans[i]); 
    return 0;
}