例题9-3 UVa 1347

/*基础dpUVa 1347 Tour时间: 2017/05/06题意：给定平面上n个点的坐标，要求找出最短路线从最左边的点走到最右边的点再回来，并满足除了最左最右两点，其他点都有且只经过一次。题解：问题转化为两个人从最左边的点到最右边的点路线长总和最短，并满足两个人的路线互相不能经过同一个点，但总的经过所有点。dp[i][j] 代表前i个点都走过，其中一个人走得远点的人走到第i个点，另一个人走到第j个点的最短路线。(即i > j)dp[i+1][i] = dp[i][j]+dist(j,i+1);dp[i+1][j] = dp[i][j]+dist(i,i+1);*/#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<map>#include<queue>#include<cmath>#include<algorithm>#include<deque>typedef long long LL;using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")const int INF=0x3f3f3f3f;const int N = 1000+10;const double esp = 1e-6;struct Point{    double x,y;}a[N];bool cmp(Point f1,Point f2){    return f1.x < f2.x;}double dist(Point f1,Point f2){    return sqrt((f1.x-f2.x)*(f1.x-f2.x)+(f1.y-f2.y)*(f1.y-f2.y));}double dp[N][N];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                dp[i][j] = 1000000000;        for(int i = 1; i <= n; i++)            scanf("%lf%lf",&a[i].x,&a[i].y);        sort(a+1,a+n+1,cmp);        dp[1][1] = 0;        for(int i = 2; i <= n; i++)        {            dp[i][i-1] = dp[i-1][1]+dist(a[i],a[1]);            for(int j = 1; j < i; j++)            {                dp[i][i-1] = min(dp[i][i-1],dp[i-1][j]+dist(a[i],a[j]));                if(j != i-1)                    dp[i][j] = dp[i-1][j]+dist(a[i],a[i-1]);            }        }        double ans = 1000000000;        for(int i = 1; i < n; i++)            ans = min(ans,dp[n][i]+dist(a[i],a[n]));        printf("%.2lf\n",ans);    }    return 0;}