原题地址：http://poj.org/problem?id=2576
Tug of War
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Time Limit: 3000MS |
Memory Limit: 65536K |
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Total Submissions: 8525 |
Accepted: 2320 |
Description
A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.
Input
The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.
Output
Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.
Sample Input
3
100
90
200
Sample Output
190 200
题意：有n个人要分成两队进行拔河比赛，要求两队人数之差不超过1，且体重之和尽可能接近。
思路：先按体重从小到大排序，然后把前一半划为a队，后一半划为b队，算出体重之差作为当前最小值。生成两个随机数x,y，作为进行交换的队员编号，即把a队的x号队员放到b队，把b队的y号队员放到a队，检查交换后体重之差是否更小，若是则交换过来并记录新的最小值，否则不交换。重复操作N次，只要RP足够好即可AC。
1 #include <cstdio>
2 #include <algorithm>
3 #include <stdlib.h>
4 #include <time.h>
5 using namespace std;
6
7 int main(void)
8 {
9 srand((unsigned)time(NULL));
10 int N, n, a[110], b[55], c[55], nb, nc, sumb, sumc, close, x, y, sx, sy;
11 while(~scanf("%d", &n))
12 {
13 for(int i=0; i<n; ++i)
14 scanf("%d", a+i);
15 if(n==1)
16 {
17 printf("0 %d\n", a[0]);
18 continue;
19 }
20 sort(a, a+n);
21 sumb = sumc = 0;
22 if(n&1)
23 {
24 nb = n/2+1;
25 nc = n - nb;
26 int i = 0;
27 for(; i<nb; ++i)
28 b[i] = a[i], sumb += a[i];
29 for(; i<n; ++i)
30 c[i-nb] = a[i], sumc += a[i];
31 }
32 else
33 {
34 nb = nc = n/2;
35 int i = 0;
36 for(; i<nb; ++i)
37 b[i] = a[i], sumb += a[i];
38 for(; i<n; ++i)
39 c[i-nb] = a[i], sumc += a[i];
40 }
41 close = abs(sumb-sumc);
42 N = 1000000;
43 while(N--)
44 {
45 x = rand() % nb;
46 y = rand() % nc;
47 sx = sumb - b[x] + c[y];
48 sy = sumc + b[x] - c[y];
49 if(abs(sx-sy) < close)
50 {
51 sumb = sx;
52 sumc = sy;
53 b[x] ^= c[y], c[y] ^= b[x], b[x] ^= c[y];
54 close = abs(sx-sy);
55 }
56 }
57 if(sumb > sumc)
58 sumb ^= sumc, sumc ^= sumb, sumb ^= sumc;
59 printf("%d %d\n", sumb, sumc);
60 }
61 }