Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Wen Chen is the captain of a rescue boat. One of his important tasks is to visit a group of islands once a day to check if everything is all right. Captain Wen starts from the west-most island, makes a pass to the east-most island visiting some of the islands, then makes a second pass from the east-most island back to the first one visiting the remaining islands. In each pass Captain Wen moves steadily east (in the first pass) or west (in the second pass), but moves as far north or south as he needs to reach the islands. The only complication is that there are two special islands where Wen gets fuel for his boat, so he must visit them in separate passes. Figure 7 shows the two special islands in pink (1 and 3) and one possible path Captain Wen could take.

Calculate the length of the shortest path to visit all the islands in two passes when each island's location and the identification of the two special islands are given.

Input 

The input consists of multiple test cases. The data for each case begins with a line containing 3 integers n (4n100), b₁, and b₂ (0 < b₁, b₂ < n - 1 and b₁ b₂), where n is the number of islands (numbered 0 to n - 1) and b₁ and b₂ are the two special islands. Following this, there are n lines containing the integer x - and y -coordinates of each island (0x, y2000), starting with island 0. No two islands have the same x -coordinate and they are in order from west-most to east-most (that is, minimum x -coordinate to maximum x -coordinate).

Input for the last case is followed by a line containing 3 zeroes.

Output 

For each case, display two lines. The first line contains the case number and the length of the shortest tour Captain Wen can take to visit all the islands, rounded and displayed to the nearest hundredth. The second line contains a space-separated list of the islands in the order that they should be visited, starting with island 0 and 1, and ending with island 0. Each test case will have a unique solution. Follow the format in the sample output.

Sample Input 

5 1 3 
1 3 
3 4 
4 1 
7 5 
8 3 
5 3 2 
0 10 
3 14 
4 7 
7 10 
8 12 
0 0 0

Sample Output 

Case 1: 18.18 
0 1 4 3 2 0 
Case 2: 24.30 
0 1 3 4 2 0

---

題目描述：
從最左邊的點出發，第一趟必須全部往右邊走，到達最右邊的點，第二趟全部往左走，回到原點。
兩趟要走過所有點。題目要求有兩個特殊點，兩個點必須在不同趟中被走過，不可在同一趟。

再者是輸出的時候，以 0 1 開頭的方式輸出，且只有唯一解

題目解法：
雖然題目要求全往右之後，全往左。但事實上可當作兩條路徑同時向右經過左所有點。
dp[i][j] 表示第一條路徑的最後一點 i, 第二條路徑的最後一點 j, 而下一個必須被加入的點一定是
max(i, j)+1，然後考慮這一點要接在第一條路徑之後，還是第二條路徑之後。

特別規定特殊點 b1 只能在第一條，b2 只能在第二條。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
double dp[105][105];
pair<int, int> dparg[105][105];
int main() {
    int n, b1, b2, cases = 0;
    int i, j, k;
    double x[105], y[105];
    while(scanf("%d %d %d", &n, &b1, &b2) == 3) {
        if(n == 0)  break;
        for(i = 0; i < n; i++)
            scanf("%lf %lf", &x[i], &y[i]);
        double ret = 1e+60;
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++)
                dp[i][j] = 1e+60;
        }
        dp[0][0] = 0;
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                k = max(i, j)+1;
                if(k != b2) {
                    if(dp[i][k] > dp[i][j] + hypot(x[k]-x[j], y[k]-y[j])) {
                        dp[i][k] = dp[i][j] + hypot(x[k]-x[j], y[k]-y[j]);
                        dparg[i][k] = make_pair(i, j);
                    }
                }
                if(k != b1) {
                    if(dp[k][j] > dp[i][j] + hypot(x[k]-x[i], y[k]-y[i])) {
                        dp[k][j] = dp[i][j] + hypot(x[k]-x[i], y[k]-y[i]);
                        dparg[k][j] = make_pair(i, j);
                    }
                }
            }
        }
        for(i = 0; i < n; i++) {
            ret = min(ret, dp[i][n-1] + hypot(x[n-1]-x[i], y[n-1]-y[i]));
            ret = min(ret, dp[n-1][i] + hypot(x[n-1]-x[i], y[n-1]-y[i]));
        }
        printf("Case %d: %.2lf\n", ++cases, ret);
        pair<int, int> pre;
        for(i = 0; i < n; i++) {
            if(fabs(dp[i][n-1] + hypot(x[n-1]-x[i], y[n-1]-y[i])-ret) < 1e-8) {
                pre = make_pair(i, n-1);
                break;
            }
            if(fabs(dp[n-1][i] + hypot(x[n-1]-x[i], y[n-1]-y[i])-ret) < 1e-8) {
                pre = make_pair(n-1, i);
                break;
            }
        }
        int path1[105], path2[105], p1 = 0, p2 = 0;
        while(true) {
            i = pre.first, j = pre.second;
            pre = dparg[i][j];
            if(i != pre.first)
                path1[p1++] = i;
            else
                path2[p2++] = j;
            if(pre.first == 0 && pre.second == 0) {
                break;
            }
        }
        printf("0 ");
        if(path2[p2-1] == 1) {
            for(i = p2-1; i >= 0; i--)
                printf("%d ", path2[i]);
            for(i = 0; i < p1; i++)
                printf("%d ", path1[i]);
        } else {
            for(i = p1-1; i >= 0; i--)
                printf("%d ", path1[i]);
            for(i = 0; i < p2; i++)
                printf("%d ", path2[i]);
        }
        puts("0");
    }
    return 0;
}