【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 = 0, s = 1;
    char c;
    while ((c = getchar()) == ' ' || c == '\n');
    if (c == '-') s = -s; else value = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        value = value * 10 + c - 48;
    a = s * value;
}

vector<pair<double, double> > P;
int n;
double x, y, dis[205][205], dp[205][205];

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, 0, sizeof (dis));
        memset (dp , 0, sizeof (dp));
        for (int i = 0; i < n; ++ i) {
            scanf ("%lf %lf", &x, &y);
            P.push_back (make_pair(x, y));
        }
        for (int i = 0; i < P.size(); ++ i) {
            for (int j = 0; j < P.size(); ++ j) {
                dis[i + 1][j + 1] = dis[j + 1][i + 1]
                    = caldis (P[i].first, P[i].second, P[j].first, P[j].second);
            }
        }

        dp[1][2] = dis[1][2];
        //cout << dp[1][2] << endl;
        for (int j = 3; j <= n; ++ j) {
            for (int i = 1; i <= j - 2; ++ i) {
                dp[i][j] = dp[i][j - 1] + dis[j - 1][j];
            }

            dp[j - 1][j] = UINT_MAX;

            for (int k = 1; k <= j - 2; ++ k) {
                dp[j - 1][j] = min (dp[j - 1][j], dp[k][j - 1] + dis[k][j]);
            }
        }
        dp[n][n] = dp[n - 1][n] + dis[n - 1][n];
        printf ("%.2lf\n", dp[n][n]);
    }
    return 0;
}