【HDU2224】The shortest path（双调欧几里得dp）

算法导论上一道dp，挺有趣的。于是就研究了一阵。

dp(i, j)代表从左边第一个点到第i个点与从从左边最后一个点（即为第一个点）到j点的最优距离和。于是找到了子状态。

决策过程 dp[i][j] = min{dp[i-1][j] + Dis(i, i - 1), dp[i - 1][k] + Dis(k, i)} 即可。代表意思是选择最优的路径加入到回路中的去途或者归途中。

一下是代码。

 /****************************************/
 /*****            Desgard_Duan        *****/
 /****************************************/
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <stack>
 #include <map>
 #include <queue>
 #include <vector>
 #include <set>
 #include <functional>
 #include <cmath>
 #include <numeric>
 #include <limits.h>

 using namespace std;

 inline void get_val(int &a) {
     int value = , s = ;
     char c;
     while ((c = getchar()) == ' ' || c == '\n');
     if (c == '-') s = -s; else value = c - ;
     while ((c = getchar()) >= '' && c <= '')
         value = value *  + c - ;
     a = s * value;
 }

 vector<pair<double, double> > P;
 int n;
 double x, y, dis[][], dp[][];

 double caldis (double x1, double y1, double x2, double y2) {
     return sqrt ((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
 }

 int main () {
     while (~scanf ("%d", &n)) {
         P.clear();
         memset (dis, , sizeof (dis));
         memset (dp , , sizeof (dp));
         for (int i = ; i < n; ++ i) {
             scanf ("%lf %lf", &x, &y);
             P.push_back (make_pair(x, y));
         }
         for (int i = ; i < P.size(); ++ i) {
             for (int j = ; j < P.size(); ++ j) {
                 dis[i + ][j + ] = dis[j + ][i + ]
                     = caldis (P[i].first, P[i].second, P[j].first, P[j].second);
             }
         }

         dp[][] = dis[][];
         //cout << dp[1][2] << endl;
         for (int j = ; j <= n; ++ j) {
             for (int i = ; i <= j - ; ++ i) {
                 dp[i][j] = dp[i][j - ] + dis[j - ][j];
             }

             dp[j - ][j] = UINT_MAX;

             for (int k = ; k <= j - ; ++ k) {
                 dp[j - ][j] = min (dp[j - ][j], dp[k][j - ] + dis[k][j]);
             }
         }
         dp[n][n] = dp[n - ][n] + dis[n - ][n];
         printf ("%.2lf\n", dp[n][n]);
     }
     return ;
 }