51nod 1021 石子归并(dp)

51nod 1021 石子归并

题解：从i到j合并的最小值：dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + sum[j] - sum[i-1]); 最后dp[1][n]即为所求结果。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define CLR(a,b) memset((a),(b),sizeof((a)))
 using namespace std;

 const int inf = 0x3f3f3f3f;
 const int N = ;
 int n;
 int a[N];
 int dp[N][N];//i到j合并的最小值
 int sum[N];
 int main(){
     int t, i, j, k, len;
     sum[] = ;
     scanf("%d", &n);
     for(i = ; i <= n; ++i){
         scanf("%d", &a[i]);
         sum[i] = sum[i - ] + a[i];
     }
     for(i =  ;i <= n; ++i)
         dp[i][i] = ;
     for(len = ; len <= n; ++len){//合并的长度
         for(i = ; i <= n - len + ; ++i){//起点
             j = i + len - ;//终点
             dp[i][j] = inf;
             for(k = i; k < j; ++k){
                 dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+][j] + sum[j] - sum[i-]);
             }
         }
     }
     printf("%d\n", dp[][n]);
     return ;
 }