hdu_2224_The shortest path(dp)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2224

题意：双调欧几里德旅行商经典问题，找一条最短回路使得该路经过所有的点

题解：dp[i][j]=dp[i-1][j]+dis(i,i-1),dp[i][i-1]=Min(dp[i][i-1],dp[i-1][j]+dis(i,j));，注意这里题目的数据给的是从左往右的，所以不需要排序

#include<cstdio>
#include<cmath>
#define FFC(i,a,b) for(int i=a;i<=b;i++)
const int maxn=210;
double dp[maxn][maxn],inf=1e9;
struct node{double x,y;}g[maxn];
double Min(double a,double b){return a>b?b:a;}
double dis(int i,int j){return sqrt((g[i].x-g[j].x)*(g[i].x-g[j].x)+(g[i].y-g[j].y)*(g[i].y-g[j].y));}
double fuck(int n){
    FFC(i,1,n)dp[i][i-1]=inf;
    dp[1][1]=0,dp[2][1]=dis(2,1);
    FFC(i,3,n)FFC(j,1,i-2)
    dp[i][j]=dp[i-1][j]+dis(i,i-1),dp[i][i-1]=Min(dp[i][i-1],dp[i-1][j]+dis(i,j));
    return dp[n][n-1]+dis(n,n-1);
}
int main(){
    int n;
    while(~scanf("%d",&n)){
        FFC(i,1,n)scanf("%lf%lf",&g[i].x,&g[i].y);
        printf("%.2lf
",fuck(n));
    }
    return 0;
}