dp四边形优化 Hdu 3516 Tree Construction 题解

累加器传送门：

http://blog.csdn.net/NOIAu/article/details/71775000

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题目传送门：

https://vjudge.net/problem/HDU-3516

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题目：

Consider a two-dimensional space with a set of points (xi, yi) that satisfy xi < xj and yi > yj for all i < j. We want to have them all connected by a directed tree whose edges go toward either right (x positive) or upward (y positive). The figure below shows an example tree.

Write a program that finds a tree connecting all given points with the shortest total length of edges.

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输入：

The input begins with a line that contains an integer n (1 <= n <= 1000), the number of points. Then n lines follow. The i-th line contains two integers xi and yi (0 <= xi, yi <= 10000), which give the coordinates of the i-th point.

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输出：

Print the total length of edges in a line.

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样例输入：

5
1 5
2 4
3 3
4 2
5 1
1
10000 0

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样例输出：

12
0

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题目大意：

二维平面上有n个，这些点满足
i < j时，xi < xj, yi > yj
只用朝x轴或y轴正方向的边连起来，形成一棵树，使得所有边的长度最小
其中 n <= 1000

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很容易想到一个dp转移方程dp[i][j]=min{dp[i][k]+dp[k+1][j]+cost[i][k][k+1][j]}
这个cost[i][k][k+1][j]没必要预处理出来
模拟一下可以得到
它的值等于p[k].y-p[j].y+p[k+1].x-p[i].x
所以四边形优化就可以了，这篇博文有四边形优化详细讲解
http://blog.csdn.net/NOIAu/article/details/72514812

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代码如下

#include<cstdio>#include<cstring>#include<iostream>#define MAXN 1000+10using namespace std;int n;int s[MAXN][MAXN],dp[MAXN][MAXN];struct P{    int x,y;}p[MAXN];bool init(){    if(scanf("%d",&n)==EOF) return false;    for(register int i=1;i<=n;i++) scanf("%d%d",&p[i].x,&p[i].y);    for(register int i=1;i<=n;i++){        dp[i][i]=0;        s[i][i]=i;    }    return true;}void dpp(){    for(register int i=n;i>=1;i--){        for(register int j=i+1;j<=n;j++){            int temp=0x7fffffff;            int te;            for(int k=s[i][j-1];k<=s[i+1][j];k++){                if(k+1<=j){                    if(temp>dp[i][k]+dp[k+1][j]+p[k].y-p[j].y+p[k+1].x-p[i].x){                        temp=dp[i][k]+dp[k+1][j]+p[k].y-p[j].y+p[k+1].x-p[i].x;                        te=k;                    }                }            }            dp[i][j]=temp;            s[i][j]=te;        }    }    cout<<dp[1][n]<<endl;}int main(){    while(init()){        dpp();    }    return 0;}