标签：
述 有N堆石子排成一排，每堆石子有一定的数量。现要将N堆石子并成为一堆。合并的过程只能每次将相邻的两堆石子堆成一堆，每次合并花费的代价为这两堆石子的和，经过N-1次合并后成为一堆。求出总的代价最小值。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAX = 1000; const int inf = 9999999; int dp[MAX][MAX],ans[MAX],sum[MAX]; int main() { int n; scanf("%d",&n); memset(dp,0,sizeof(dp)); sum[0] = 0; for(int i = 1; i <= n ; i++){ scanf("%d",&ans[i]); sum[i] = sum[i-1] + ans[i]; } int k,i,j; for(int k = 2; k <= n ; k++){ for(int i = 1; i <= n - k + 1;i++){ j = i + k - 1; dp[i][j] = inf; for(int m = i; m < j ; m++){ dp[i][j] = min(dp[i][j],dp[i][m]+dp[m+1][j]+sum[j]-sum[i-1]); } } } printf("%d\n",dp[1][n]); return 0; }
可以用平行四边形优化，即用一个数组s来存储每个最优位置。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAX = 1000; const int inf = 9999999; int dp[MAX][MAX],ans[MAX],sum[MAX],s[MAX][MAX]; int main() { int n; scanf("%d",&n); memset(dp,0,sizeof(dp)); sum[0] = 0; for(int i = 1; i <= n ; i++){ scanf("%d",&ans[i]); sum[i] = sum[i-1] + ans[i]; s[i][i] = i; } int k,i,j; for(int k = 2; k <= n ; k++){ for(int i = 1; i <= n - k + 1;i++){ j = i + k - 1; dp[i][j] = inf; for(int m = s[i][j-1]; m <= s[i+1][j];m++){ if(dp[i][j] > dp[i][m]+dp[m+1][j]+sum[j]-sum[i-1]){ dp[i][j] = dp[i][m]+dp[m+1][j]+sum[j]-sum[i-1]; s[i][j] = m; } } } } printf("%d\n",dp[1][n]); return 0; }
标签：
原文地址：http://www.cnblogs.com/zero-begin/p/4345143.html
