UVa Problem Solution: 10037 – Bridge

Given the two fastest people as f1 and f2, f1 < f2, and the two slowest people as s1 and s2, s1 < s2. We want to use the two fastest people to bring the two slowest people to cross the bridge. There are two efficient ways of doing this:
  1) f1, f2 go --> f1 back --> s1, s2 go --> f2 back: time1 = f2 + f1 + s2 + f2
  2) f1, s1 go --> f1 back --> f1, s2 go --> f1 back: time2 = s1 + f1 + s2 + f1
Then, we can compare time1 with time2 to choose the better way.

Code:

1. /*************************************************************************
2.  * Copyright (C) 2008 by liukaipeng                                      *
3.  * liukaipeng at gmail dot com                                           *
4.  *************************************************************************/
6. /* @JUDGE_ID 00000 10037 C++ "Bridge" */
8. #include <algorithm>
9. #include <deque>
10. #include <fstream>
11. #include <iostream>
12. #include <list>
13. #include <map>
14. #include <queue>
15. #include <set>
16. #include <stack>
17. #include <string>
18. #include <vector>
20. using namespace std;
21.      
22. int const timecount = 1000;
23.      
24. void cross_bridge(int *times, int ntimes, int strategy[][2])
25. {
26.   sort(times, times + ntimes);
27.   int i = 0;
28.   for (; i < ntimes / 2 - 1; ++i) {
29.     int time1 = times[1] + times[0] + times[ntimes-2*i-1] + times[1];
30.     int time2 = times[ntimes-2*i-1] + times[0] + times[ntimes-2*i-2] + times[0];
31.     if (time1 < time2) {
32.       strategy[4*i+1][0] = times[0];
33.       strategy[4*i+1][1] = times[1];
34.       strategy[4*i+2][0] = times[0];
35.       strategy[4*i+3][0] = times[ntimes-2*i-2];
36.       strategy[4*i+3][1] = times[ntimes-2*i-1];
37.       strategy[4*i+4][0] = times[1];
38.       strategy[0][0] += time1;
39.     } else {
40.       strategy[4*i+1][0] = times[0];
41.       strategy[4*i+1][1] = times[ntimes-2*i-1];
42.       strategy[4*i+2][0] = times[0];
43.       strategy[4*i+3][0] = times[0];
44.       strategy[4*i+3][1] = times[ntimes-2*i-2];
45.       strategy[4*i+4][0] = times[0];
46.       strategy[0][0] += time2;
47.     }
48.   }
49.   strategy[4*i+1][0] = times[0];
50.   strategy[4*i+1][1] = times[1];
51.   strategy[0][0] += times[1];
52.   if (ntimes % 2 == 1) {
53.     strategy[4*i+2][0] = times[0];
54.     strategy[4*i+3][0] = times[0];
55.     strategy[4*i+3][1] = times[2];
56.     strategy[0][0] += times[0] + times[2];
57.   }
58. }
59.      
60. int main(int argc, char *argv[])
61. {
62. #ifndef ONLINE_JUDGE
63.   filebuf in, out;
64.   cin.rdbuf(in.open((string(argv[0]) + ".in").c_str(), ios_base::in));
65.   cout.rdbuf(out.open((string(argv[0]) + ".out").c_str(), ios_base::out));
66. #endif
68.   int ncases;
69.   cin >> ncases;
70.   while (ncases-- > 0) {
71.     int ntimes;
72.     cin >> ntimes;
73.     int times[timecount];
74.     for (int i = 0; i < ntimes; ++i) cin >> times[i];
75.     if (ntimes < 2) {
76.       cout << times[0] << '/n' << times[0] << '/n';
77.     } else {
78.       int strategy[(timecount-1)*2][2] = {{0}};
79.       cross_bridge(times, ntimes, strategy);
80.       for (int i = 0; i < (ntimes - 1) * 2; ++i) {
81.         cout << strategy[i][0];
82.         if (i % 2 == 1) cout << ' ' << strategy[i][1];
83.         cout << '/n';
84.       }
85.     }
86.     if (ncases > 0) cout << '/n';
87.   }
89.   return 0;
90. }
91.