UVA 1347 Tour

动态规划

题意可以理解为两个人同时从最左点出发，沿着两条不同的路径走到最右点（除了起点和终点每个点走且仅走一次）

状态 dp[i][j]指当前两人分别走到i，j点。且设i>j；

则有：dp[i+1][i]=min (dp[i+1][i],dp[i][j]+dist[i][i+1]);

　　   dp[i+1][j]=min (dp[i+1][j],dp[i][j]+dist[j][i+1]);

因为每个点都要走到，所以每次走到没走到的第一个点。

最终结果是 dp[n-1][i]+dist[i][n]+dist[n-1][n];(0<i<n-1) 最终点要走两次。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const double inf=99999999.0;
    int n;
    double a[1005][2];
    double dist[1005][1005];
    double dp[1005][1005];

double getdist (double x1,double y1,double x2,double y2){
    double x,y;
    x=fabs (x1-x2);
    y=fabs (y1-y2);
    return sqrt (x*x+y*y);
}    

int main (){//cout<<inf<<endl;
    while (~scanf ("%d",&n)){
        for (int i=0;i<n;i++){
            scanf ("%lf%lf",&a[i][0],&a[i][1]);
        }
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                dist[i][j]=getdist (a[i][0],a[i][1],a[j][0],a[j][1]);//cout<<i<<" "<<j<<":"<<dist[i][j]<<"    ";
        for (int i=0;i<=n;i++)
            for (int j=0;j<=n;j++)
                dp[i][j]=inf;
        dp[0][0]=0;
        for (int i=0;i<n-2;i++){
            for (int j=0;(i==0&&j==0)||j<i;j++){
                dp[i+1][j]=min (dp[i+1][j],dp[i][j]+dist[i][i+1]);
                dp[i+1][i]=min (dp[i+1][i],dp[i][j]+dist[j][i+1]);//cout<<dp[i+1][j]<<" "<<dp[i+1][i]<<endl;
            }
        }//cout<<endl;
        double ans=inf;
        for (int i=0;i<n-2;i++)
            ans=min (ans,dp[n-2][i]+dist[n-1][n-2]+dist[n-1][i]);//cout<<n-2<<" "<<i<<endl,cout<<dp[n-2][i]+dist[n-1][n-2]+dist[n-1][i]<<endl;
        printf ("%.2f
",ans);
    }
    return 0;
}