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 = 101;
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[0] = 0;
    scanf("%d", &n);
    for(i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
        sum[i] = sum[i - 1] + a[i];
    }
    for(i = 1 ;i <= n; ++i)
        dp[i][i] = 0;
    for(len = 2; len <= n; ++len){//合并的长度
        for(i = 1; i <= n - len + 1; ++i){//起点
            j = i + len - 1;//终点
            dp[i][j] = inf;
            for(k = i; k < j; ++k){
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + sum[j] - sum[i-1]);
            }
        }
    }
    printf("%d\n", dp[1][n]);
    return 0;
}