Codeforces 1102F Elongated Matrix 状压dp

Elongated Matrix

预处理一下两两之间的最小值， 然后直接dp。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 1e4 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

int n, m;
int Mat[][N];
int go[][];
int go2[][];
int dp[ << ][];

int main() {
    scanf("%d%d", &n, &m);
    for(int i = ; i < n; i++)
        for(int j = ; j < m; j++)
            scanf("%d", &Mat[i][j]);
    int ans = ;
    if(n == ) {
        ans = inf;
        for(int i = ; i < m - ; i++) ans = min(ans, abs(Mat[][i] - Mat[][i + ]));
        printf("%d\n", ans);
        return ;
    }
    for(int i = ; i < n; i++) {
        for(int j = ; j < n; j++) {
            if(i == j) continue;
            go[i][j] = go2[i][j] = inf;
            for(int k = ; k < m; k++)
                    go[i][j] = min(go[i][j], abs(Mat[i][k] - Mat[j][k]));
            for(int k = ; k < m - ; k++) {
                    go2[i][j] = min(go2[i][j], abs(Mat[i][k] - Mat[j][k + ]));
            }
        }
    }
    for(int start = ; start < n; start++) {
        memset(dp, -, sizeof(dp));
        dp[ << start][start] = inf;
        for(int S = ; S < ( << n); S++) {
            for(int i = ; i < n; i++) {
                if(dp[S][i] < ) continue;
                for(int j = ; j < n; j++) {
                    if(S >> j & ) continue;
                    dp[S|(<<j)][j] = max(dp[S|(<<j)][j], min(go[i][j], dp[S][i]));
                }
            }
        }
        for(int i = ; i < n; i++)
                ans = max(ans, min(go2[i][start], dp[(<<n)-][i]));
    }
    printf("%d\n", ans);
    return ;
}

/*
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