hdu3506(dp四边形优化)

来源：互联网发布：cortana是什么软件编辑：程序博客网时间：2024/05/21 12:39

链接：点击打开链接

题意：求环形石子合并

代码：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;int s[2005][2005],dp[2005][2005],sum[2005][2005];int main(){                                     //区间dp    int n,i,j,k,l,ans;                          //m(i,j)=min{(m(i,k-1)+m(k,j)+w(i,j)}(i<k<=j)    while(scanf("%d",&n)!=EOF){                 //当i<i'<j<j'并且        memset(dp,INF,sizeof(dp));              //w(i,j)+w(i',j')≤w(i',j)+w(j,i')时        for(i=1;i<=n;i++){                      //函数m也满足这个关系，从而可以用四边形优化            scanf("%d",&sum[i][i]);             //也就是k的范围在s(i,j+1)~s(i+1,j)内            sum[i+n][i+n]=sum[i][i];            //s(i,j)代表函数m(i,j)对应的决策变量的最大值            dp[i][i]=dp[i+n][i+n]=0;            //具体证明看论文:            s[i][i]=i,s[i+n][i+n]=i+n;          //http://wenku.baidu.com/view/be418243a8956bec0975e3bf.html?re=view        }        for(i=1;i<=2*n;i++)        for(j=i+1;j<=2*n;j++)        sum[i][j]=sum[i][j-1]+sum[j][j];        for(l=2;l<=2*n;l++){            for(i=1;i<=2*n-l+1;i++){                j=i+l-1;                for(k=s[i][j-1];k<=s[i+1][j];k++){                    if(dp[i][j]>dp[i][k]+dp[k+1][j]+sum[i][j]){                        dp[i][j]=dp[i][k]+dp[k+1][j]+sum[i][j];                        s[i][j]=k;                    }                }            }        }        ans=INF;        for(i=1;i<=n;i++)        ans=min(ans,dp[i][i+n-1]);        printf("%d\n",ans);    }    return 0;}

1 0