UVa Problem 10037 Bridge （過橋）

// Bridge （過橋）// PC/UVa IDs: 110403/10037, Popularity: B, Success rate: low Level: 3// Verdict: Accepted// Submission Date: 2011-05-22// UVa Run Time: 0.012s//// 著作權（C）2011，邱秋。metaphysis # yeah dot net//// 首先將所有過橋時間排序，以下為獲得最小過橋時間的演算法：// （1）過橋總人數為 1，該人的過橋時間即為最短過橋時間。// （2）過橋總人數為 2，過橋時間較大的人的過橋時間即為最短過橋時間。// （3）過橋總人數為 3，假設為A，B，C，則（AB），（A），（AC）策略和 （AC），（A），（AB）策略// 的時間相同，3人過橋時間之和為最短時間。// （4）過橋總人數大於3，假設最前面為A，B兩人，最後為Y，Z兩人，有兩種策略：（AB），（A），（YZ），// （B），（AB）和（AZ），（A），（AY），（A），（AB）。比較兩種策略那種過橋時間少就選那種，然後// 將總人數減去 2，若總人數仍大於 3，繼續該步驟直到剩下需要過橋的人數小於等於 3。可用遞迴或直接迭代實現。#include <iostream>#include <algorithm>using namespace std;#define MAXSIZE (1000 + 1)int shortest_time(int time[], int capacity){if (capacity == 1)return time[0];if (capacity == 2)return time[1];if (capacity == 3)return time[0] + time[1] + time[2];if (2 * time[1] < (time[0] + time[capacity - 2]))return time[0] + 2 * time[1] + time[capacity - 1] + shortest_time(time, capacity - 2);elsereturn 2 * time[0] + time[capacity - 2] + time[capacity - 1] + shortest_time(time, capacity - 2);}void bridge(int time[], int capacity){if (capacity == 1){cout << time[0] << "\n";return;}if (capacity == 2){cout << time[0] << " " << time[1] << "\n";return;}if (capacity == 3){cout << time[0] << " " << time[2] << "\n";cout << time[0] << "\n";cout << time[0] << " " << time[1] << "\n";return;}if (2 * time[1] < (time[0] + time[capacity - 2])){cout << time[0] << " " << time[1] << "\n";cout << time[0] << "\n";cout << time[capacity - 2] << " " << time[capacity - 1] << "\n";cout << time[1] << "\n";}else{cout << time[0] << " " << time[capacity - 1] << "\n";cout << time[0] << "\n";cout << time[0] << " " << time[capacity - 2] << "\n";cout << time[0] << "\n";}bridge(time, capacity - 2);}int main(int ac, char *av[]){int cases;int time[MAXSIZE];int capacity, index;cin >> cases;while (cases--){// 總人數。cin >> capacity;// 讀取每個人的過橋時間。index = 0;while (index < capacity)cin >> time[index++];// 將用時數組予以排序。sort(time, time + capacity);// 計算最短過橋時間，輸出過橋順序。cout << shortest_time(time, capacity) << endl;bridge(time, capacity);if (cases)cout << endl;}return 0;}