洛谷 P1880 [NOI1995]石子合并（区间DP）

嗯...

题目链接：https://www.luogu.org/problem/P1880

这道题特点在于石子是一个环，所以让a[i+n] = a[i]（两倍长度）即可解决环的问题，然后注意求区间最小值的时候dp要初始化为一个很大的数...

AC代码：

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 
 using namespace std;
 
 int dp1[][], dp2[][], sum[], a[];
 
 int main(){
     int n;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) {scanf("%d", &a[i]); a[i + n] = a[i];}
     for(int i = ; i <= n * ; i++) {sum[i] = sum[i - ] + a[i];}
     for(int l = ; l <= n; l++){
         for(int i = ; i <=  * n; i++){
             int j = i + l;
             dp2[i][j] = 0x3f3f3f;// 求最小值初始化为很大
             if(j >  * n) break;
             for(int k = i; k < j; k++){
                 dp1[i][j] = max(dp1[i][j], dp1[i][k] + dp1[k + ][j] + sum[j] - sum[i - ]);
                 dp2[i][j] = min(dp2[i][j], dp2[i][k] + dp2[k + ][j] + sum[j] - sum[i - ]);
             }
         }
     }
     int maxx = , minn = 0x3f3f;
     for(int i = ; i <= n; i++){
         int j = i + n - ;
         maxx = max(maxx, dp1[i][j]);
         minn = min(minn, dp2[i][j]);
     }
     printf("%d\n%d\n", minn, maxx);
     return ;
 }

AC代码