hdu3516 Tree Construction (四边形不等式)

题意：给定一些点(xi,yi)(xj,yj)满足:i<j,xi<xj,yi>yj。用下面的连起来，使得所有边的长度最小？

题解：直接给出吧

f[i][j]=min(f[i][k]+f[k+1][j]+cost(i,j)

cost(i,j)=a[k].y-a[j].y+a[k+1].x-a[i].x;

明显了吧

 

证明一下，搞一搞，四边形性质就出来了，模板题吧。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#define N 1007
#define M 500007
#define inf 2000000009
using namespace std;

int n;
int f[N][N],s[N][N];
struct Node{int x,y;}a[M];

int cost(int i,int j,int k)
{
    if (k>=j) return inf; 
    return a[k].y-a[j].y+a[k+1].x-a[i].x;
}
int main()
{
    while (~scanf("%d",&n))
    {
        for (int i=1;i<=n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        for (int i=1;i<=n;i++)
            s[i][i]=i;
        memset(f,0,sizeof f);
        for (int L=2;L<=n;L++)
            for (int i=1;i+L-1<=n;i++)
             {
                int j=L+i-1;f[i][j]=inf;
                for (int k=s[i][j-1];k<=s[i+1][j];k++)
                {
                    int tmp=f[i][k]+f[k+1][j]+cost(i,j,k);
                    if (tmp<f[i][j]) f[i][j]=tmp,s[i][j]=k;
                }
            }
        printf("%d
",f[1][n]); 
    }
}