UVA1347---Tour(dp,双调TSP)

$dp[i][j]$表示在1~max(i,j)都已经被走过的情况下,第一个人在i点,第二个人在j点时,走完剩下的点还需要的最短距离
规定第一个人领先第二个人
所以$dp[i][j]$可以转移到$dp[i + 1][j]$和$dp[i + 1][i]$(等价于$dp[i][i + 1]$)

/*************************************************************************
    > File Name: 平常练习/uva1347.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月25日 星期一 20时27分03秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

double dp[1100][1100];
PLL point[1100];

double dist(PLL a, PLL b) {
    int x = a.first - b.first;
    int y = a.second - b.second;
    return sqrt(x * x * 1.0 + y * y * 1.0);
}

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] = 1000000000.0;
            }
        }
        int x, y;
        dp[1][1] = 0.0;
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &x, &y);
            point[i] = make_pair(x, y);
        }
        for (int i = 1; i < n - 1; ++i) {
            for (int j = 1; j <= i; ++j) {
                if (dp[i][j] != 1000000000.0) {
                    double d = dist(point[i + 1], point[i]);
                    dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + d);
                    d = dist(point[i + 1], point[j]);
                    dp[i + 1][i] = min(dp[i + 1][i], dp[i][j] + d);
                }
            }
        }
        double ans = 1000000000.0;
        for (int i = 1; i < n - 1; ++i) {
            ans = min(ans, dp[n - 1][i] + dist(point[n - 1], point[n]) + dist(point[n], point[i]));
        }
        printf("%.2f\n", ans);
    }
    return 0;
}