hdu 2829

来源：互联网发布：企业招标软件编辑：程序博客网时间：2024/04/28 07:14

稍微比划一下，可以知道cost[i][j]满足四边形不等式的规则

#include <cstdio>#include <cstring>const int MAXN = 1000 + 123;int dp[MAXN][MAXN], cost[MAXN][MAXN], s[MAXN][MAXN];int a[MAXN], sum[MAXN];int main(){    int n, m ;    while(scanf("%d%d", &n, &m) != EOF)    {        if(n == 0 && m == 0)  break;        sum[0] = 0;        for(int i = 1; i  <= n; i++)        {            scanf("%d", &a[i]);            sum[i] = sum[i-1] + a[i];        }        memset(cost, 0, sizeof(cost));        for(int i = 1; i <= n; i++)          for(int j = i + 1; j <= n; j++)             cost[i][j] = cost[i][j-1] + (sum[j-1] - sum[i-1]) * a[j];        for(int i = 1; i <= n; i++)        {            s[i][0] = 0;            dp[i][0] = cost[1][i];        }        for(int k = 1; k <= m; k++)        {            s[n+1][k] = n - 1;            for(int i = n; i >= k + 1; i--)            {                dp[i][k] = dp[k][k-1] + cost[k+1][i];                s[i][k] = k;                for(int j = s[i][k-1]; j <= s[i+1][k]; j++)                {                    if(dp[j][k-1] + cost[j+1][i] < dp[i][k])                    {                        dp[i][k] = dp[j][k-1] + cost[j+1][i];                        s[i][k] = j;                    }                }            }        }        printf("%d\n",dp[n][m]);    }    return 0;}