Codeforces1102F Elongated Matrix 【状压DP】

题目分析：

这题瞎搞一个哈密尔顿路，对于起点不同的分开跑就可以过了。

$O(n^3*2^n)$

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

const int maxn = 17;
const int maxm = 10200;

int n,m;
int a[maxn][maxm],f[maxn][maxn],g[maxn][maxn];

int dp[maxn][(1<<16)+5],arr[maxn][(1<<16)+5];

void init(){
    for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){
        f[i][j] = g[i][j] = 1e9;
        for(int k=1;k<=m;k++)f[i][j] = min(f[i][j],abs(a[i][k]-a[j][k]));
        for(int k=1;k<m;k++) g[i][j]=min(g[i][j],abs(a[i][k+1]-a[j][k]));
    }
    }
}

queue<pair<int,int> > q;
void divide(){
    int ans = 0;
    for(int i=1;i<=n;i++){
    memset(dp,0,sizeof(dp));
    memset(arr,0,sizeof(arr));
    dp[i][1<<i-1] = 1e9; arr[i][1<<i-1] = 1;
    q.push(make_pair(i,1<<i-1));
    while(!q.empty()){
        pair<int,int> k = q.front(); q.pop();
        for(int j=1;j<=n;j++){
        if(k.second&(1<<j-1)) continue;
        dp[j][k.second|(1<<j-1)] =
            max(dp[j][k.second|(1<<j-1)],min(dp[k.first][k.second],f[k.first][j]));
        if(!arr[j][k.second|(1<<j-1)]){
            q.push(make_pair(j,k.second|(1<<j-1)));
            arr[j][k.second|(1<<j-1)] = 1;
        }
        }
    }
    for(int j=1;j<=n;j++){
        if(i == j) continue;
        ans = max(ans,min(dp[j][(1<<n)-1],g[i][j]));
    }
    }
    printf("%d
",ans);
}

void read(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
}

int main(){
    read();
    if(n == 1){
    int ans = 1e9;
    for(int i=2;i<=m;i++){ans = min(ans,abs(a[1][i]-a[1][i-1]));}
    printf("%d
",ans);
    return 0;
    }
    init();
    divide();
    return 0;
}