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：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cmath>

 4 #include <vector>

 5 #include <cstring>

 6 #include <algorithm>

 7 #include <string>

 8 #include <set>

 9 #include <ctime>

10 #include <queue>

11 #include <map>

12 #include <functional>

13 #include <numeric>

14 #include <sstream>

15 #include <stack>

16 

17 #define CL(arr, val)    memset(arr, val, sizeof(arr))

18 #define REP(i, n)       for((i) = 0; (i) < (n); ++(i))

19 #define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))

20 #define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))

21 #define L(x)    (x) << 1

22 #define R(x)    (x) << 1 | 1

23 #define MID(l, r)   (l + r) >> 1

24 #define Min(x, y)   x < y ? x : y

25 #define Max(x, y)   x < y ? y : x

26 #define E(x)    (1 << (x))

27 #define iabs(x) (x) < 0 ? -(x) : (x)

28 #define OUT(x)  printf("%I64d\n", x)

29 #define lowbit(x)   (x)&(-x)

30 #define Read()    freopen("data.in", "r", stdin)

31 #define Write()   fropen("data.out", "w", stdout);

32 

33 const int eps = 1e-8;

34 typedef long long LL;

35 const double inf = 1e15;

36 

37 using namespace std;

38 

39 const int N = 1024;

40 

41 double f[N][N];

42 double dis[N][N];

43 

44 struct node {

45     int x;

46     int y;

47     bool operator < (const node cmp) const {

48         return x == cmp.x ? y < cmp.y : x < cmp.x;

49     }

50 }p[N];

51 

52 double Dis(int i, int j) {

53     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.);

54 }

55 

56 int main() {

57 //    Read();

58     int n, i, j, k;

59     double ans;

60     while(~scanf("%d", &n)) {

61         for(i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);

62         sort(p + 1, p + n + 1);

63         for(i = 1; i <= n; ++i) {

64             for(j = 1; j <= i; ++j)  {

65                 dis[i][j] = dis[j][i] = Dis(i, j);

66             }

67         }

68         f[2][1] = dis[2][1];

69         //i >= j

70         for(i = 3; i <= n; ++i) {

71             //j < i - 1

72             for(j = 1; j < i - 1; ++j)  f[i][j] = f[i-1][j] + dis[i-1][i];

73             //j == i - 1

74             j = i - 1;

75             f[i][j] = inf;

76             for(k = 1; k < j; ++k) {

77                 f[i][j] = min(f[i][j], f[j][k] + dis[k][i]);    //note: f[x][y] = f[y][x]

78             }

79         }

80 

81         if(n == 2)  ans = f[2][1];

82         else    ans = f[n][n-1] + dis[n][n-1];

83         printf("%.2f\n", ans);

84     }

85     return 0;

86 }