bzoj 1820 dp

最普通dp要4维，因为肯定有一个在上一个的位置，所以可以变为3维，然后滚动数组优化一下。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;
 
const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
 
int n, m, cur, a[][], b[];
LL dp[][][];
 
int main() {
    scanf("%d", &n);
    for(int i = ; i <= n; i++)
        for(int j = ; j <= n; j++)
            scanf("%d", &a[i][j]);
 
    b[] = ;
    while(scanf("%d", &b[++m]) != EOF){}
    memset(dp[cur], INF, sizeof(dp[cur]));
    dp[cur][][] = dp[cur][][] = ;
 
    LL ans = INF;
    for(int i = ; i <= m; i++) {
        cur ^= ;
        memset(dp[cur], INF, sizeof(dp[cur]));
        for(int j = ; j <= n; j++) {
            for(int k = ; k <= n; k++) {
                dp[cur][j][k] = min(dp[cur][j][k], dp[cur^][j][k] + a[b[i-]][b[i]]);
                dp[cur][b[i-]][k] = min(dp[cur][b[i-]][k], dp[cur^][j][k] + a[j][b[i]]);
                dp[cur][j][b[i-]] = min(dp[cur][j][b[i-]], dp[cur^][j][k] + a[k][b[i]]);
            }
        }
    }
    for(int i = ; i <= n; i++)
        for(int j = ; j <= n; j++)
            ans = min(ans, dp[cur][i][j]);
    printf("%lld\n", ans);
    return ;
}
 
/*
*/