双调欧几里得旅行商问题解法 (POJ 2677)

对于这个解法，网上题解一大堆。不过觉得还是算导官方给的解答好些，简洁又清楚。。。

 

大概总结一下(不是翻译)=.=

 首先是排序。

然后确定状态 f[i][j] (i >= j, i == j表示的只有n这个点) 表示 从i到1严格从右往左走然后再从1严格从左往右走到j 这样两条路径的最小值，当然[1, max(i, j)]区间上所有的点都是被访问过得。

存在两种状态 :

1:  j < i - 1;

dp[i][j] = dp[i-1][j] + Dis(i, i - 1);

2: j == i - 1;

dp[i][j] = min{dp[i][j], dp[j][k] + Dis(k, i)}　　//这个dp[j][k]本来应该是dp[k][j]的，但是由于dp[x][y] = dp[y][x]所以用dp[j][k]表示更方便，因为dp[i][j]满足i > j, dp[j][k]也满足j > k。 

 

 POJ 2677&&HDU 2224代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <ctime>
#include <queue>
#include <map>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   x < y ? x : y
#define Max(x, y)   x < y ? y : x
#define E(x)    (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()    freopen("data.in", "r", stdin)
#define Write()   fropen("data.out", "w", stdout);

const int eps = 1e-8;
typedef long long LL;
const double inf = 1e15;

using namespace std;

const int N = 1024;

double f[N][N];
double dis[N][N];

struct node {
    int x;
    int y;
    bool operator < (const node cmp) const {
        return x == cmp.x ? y < cmp.y : x < cmp.x;
    }
}p[N];

double Dis(int i, int j) {
    return sqrt((p[i].x - p[j].x)*(p[i].x - p[j].x)*1. + (p[i].y - p[j].y)*(p[i].y - p[j].y)*1.);
}

int main() {
//    Read();
    int n, i, j, k;
    double ans;
    while(~scanf("%d", &n)) {
        for(i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
        sort(p + 1, p + n + 1);
        for(i = 1; i <= n; ++i) {
            for(j = 1; j <= i; ++j)  {
                dis[i][j] = dis[j][i] = Dis(i, j);
            }
        }
        f[2][1] = dis[2][1];
        //i >= j
        for(i = 3; i <= n; ++i) {
            //j < i - 1
            for(j = 1; j < i - 1; ++j)  f[i][j] = f[i-1][j] + dis[i-1][i];
            //j == i - 1
            j = i - 1;
            f[i][j] = inf;
            for(k = 1; k < j; ++k) {
                f[i][j] = min(f[i][j], f[j][k] + dis[k][i]);    //note: f[x][y] = f[y][x]
            }
        }

        if(n == 2)  ans = f[2][1];
        else    ans = f[n][n-1] + dis[n][n-1];
        printf("%.2f\n", ans);
    }
    return 0;
}