poj2677 Tour

题意：

双调欧几里得旅行商问题。

思路：

dp。定义dp[i][j](i <= j)为从点j从右向左严格按照x坐标递减顺序走到点1，之后再从点1从左向右严格按照x坐标递增的顺序走到点i，并且在此过程中经过且仅经过1到j之间所有的点1次。则

dp[1][2] = dis[1][2];

dp[i][j] = dp[i][j - 1] + dis[j - 1][j]; (i < j - 1)

dp[i][j] = min(dp[k][j - 1] + dis[k][j]). (1 <= k < j - 1, i == j - 1)

最终，dp[n][n] = dp[n - 1][n] + dis[n - 1][n].

实现：

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

const int MAXN = 50, INF = 0x3f3f3f3f;
double dp[MAXN + 5][MAXN + 5], dis[MAXN + 5][MAXN + 5];
int n, x[MAXN + 5], y[MAXN + 5];

int square(int x)
{
    return x * x;
}

void init()
{
    for (int i = 1; i <= n; i++)
    {
        for (int j = i + 1; j <= n; j++)
        {
            dis[i][j] = sqrt(square(x[i] - x[j]) + square(y[i] - y[j]));
        }
    }
}

double solve()
{
    dp[1][2] = dis[1][2];
    for (int j = 3; j <= n; j++)
    {
        // i < j - 1
        for (int i = 1; i <= j - 2; i++)
        {
            dp[i][j] = dp[i][j - 1] + dis[j - 1][j];
        }
        // i == j - 1
        dp[j - 1][j] = INF;
        for (int k = 1; k <= j - 2; k++)
        {
            dp[j - 1][j] = min(dp[j - 1][j], dp[k][j - 1] + dis[k][j]);
        }
    }
    return dp[n][n] = dp[n - 1][n] + dis[n - 1][n];
}

int main()
{
    while (cin >> n)
    {
        for (int i = 1; i <= n; i++)
        {
            cin >> x[i] >> y[i];
        }
        init();
        printf("%.2f
", solve());
    }
    return 0;
}

总结：

假设一个最优选择，然后再基于该最优选择来定义问题，是动态规划的惯用手法。