HDU 3516 Tree Construction

_(:3」∠)_

四边形优化

dp[i][j]:第i个点和第j个点合并的最小花费

cost(i,j,k)：第i和第j个点通过第k个点合并所需要的花费

容易写出：dp[i][j]=min{dp[i][k]+dp[k+1][j]+cost(i,j,k)}

易证cost函数为凸

然后四边形优化之

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iostream>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#define BUG printf("hehe
")
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;

struct Point
{
    int x,y;
}point[1010];

int dp[1010][1010];
int s[1010][1010];

int cost(int i,int j,int k)
{
    return point[k].y-point[j].y+point[k+1].x-point[i].x;
}

int main()
{
    int n;
    while(scanf("%d",&n)==1) {
        for(int i=1;i<=n;++i)
            scanf("%d%d",&point[i].x,&point[i].y);
        for(int i=1;i<=n;++i)
            s[i][i]=i;
        for(int len=1;len<n;++len) {
            for(int i=1;i<=n-len;++i) {
                int j=i+len;
                dp[i][j]=INF;
                for(int k=s[i][j-1];k<=s[i+1][j]&&k<j;++k) {  //这里k<j很关键= =、不然连样例都出不了
                    if(dp[i][j]>dp[i][k]+dp[k+1][j]+cost(i,j,k)) {
                        s[i][j]=k;
                        dp[i][j]=dp[i][k]+dp[k+1][j]+cost(i,j,k);
                    }
                }
            }
        }
        printf("%d
",dp[1][n]);
    }
}