51nod 1021 石子归并(dp)

Posted 2020-08-12 GraceSkyer

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。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define CLR(a,b) memset((a),(b),sizeof((a)))
 5 using namespace std;
 6 
 7 const int inf = 0x3f3f3f3f;
 8 const int N = 101;
 9 int n;
10 int a[N];
11 int dp[N][N];//i到j合并的最小值
12 int sum[N];
13 int main(){
14     int t, i, j, k, len;
15     sum[0] = 0;
16     scanf("%d", &n);
17     for(i = 1; i <= n; ++i){
18         scanf("%d", &a[i]);
19         sum[i] = sum[i - 1] + a[i];
20     }
21     for(i = 1 ;i <= n; ++i)
22         dp[i][i] = 0;
23     for(len = 2; len <= n; ++len){//合并的长度
24         for(i = 1; i <= n - len + 1; ++i){//起点
25             j = i + len - 1;//终点
26             dp[i][j] = inf;
27             for(k = i; k < j; ++k){
28                 dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + sum[j] - sum[i-1]);
29             }
30         }
31     }
32     printf("%d\\n", dp[1][n]);
33     return 0;
34 }

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