HDU 2224 The shortest path 双调旅行商问题

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题目来源：HDU 2224 The shortest path

题意：求从1到n 然后在从n到1的最短路去的时候经过的点的顺序必须从小到大来的时候经过的点的顺序必须从大到小并且每个点只能经过一次(1和n不算) 输出最短路的长度

思路：经典题参考了大牛http://www.cppblog.com/doer-xee/archive/2009/11/30/102296.html

#include <cstdio>#include <cstring>#include <cmath>using namespace std;const int maxn = 210;double dp[maxn][maxn];double d[maxn][maxn];struct point{    double x, y;}a[maxn];double dis(int i, int j){    return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));}int main(){    int n;    while(scanf("%d", &n) != EOF)    {        for(int i = 1; i <= n; i++)            scanf("%lf %lf", &a[i].x, &a[i].y);        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= n; j++)            {                d[i][j] = dis(i, j);            }        }        dp[1][2] = d[1][2];        for(int i = 3; i <= n; i++)        {            for(int j = 1; j < i-1; j++)                dp[j][i] = dp[j][i-1] + d[i-1][i];            dp[i-1][i] = 999999999.0;            for(int j = 1; j < i-1; j++)            {                double sum = dp[j][i-1] + d[j][i];                if(dp[i-1][i] > sum)                    dp[i-1][i] = sum;            }        }        dp[n][n] = dp[n-1][n] + d[n-1][n];        printf("%.2f\n", dp[n][n]);    }    return 0;}

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