标签：lin void title ems pac git include flag http
For each test case, the input is as follows:
? One line containing the number of teams n, where n is a power of two and 2 ≤ n ≤ 1024. Teams are numbered from 1 to n, where team 1 is your favourite team.
? n lines, each containing a string of n binary digits. The k-th digit on the j-th line is ‘1’ if team j would certainly win from team k, otherwise it is ‘0’. A team cannot play against itself, therefore the j-th digit on the j-th line is ‘0’. If j ?= k, the k-th digit on the j-th line is di?erent from the j-th digit on the k-th line.
1 3
2 4
1 2
1 5
3 7
4 8
2 6
1 3
4 2
1 4
题解：这个题的构造比较难，我是想不到，构造出来之后的递归就相对比较简单了。构造的方式分为四个阶段：
1、把满足条件的队伍A和B配对，其中1打不过A，1能打过B，并且B能打过A.
2、把1和剩下的它能打过的队伍配对.
3、把1打不过的队伍相互配对.
4、把剩下的队伍配对.
能够证明按照这样的策略打过一轮之后，剩下的队伍还满足初始条件，因此可以递归求解。（构造太巧妙orz）
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 1050; 6 7 int n; 8 char gra[maxn][maxn]; 9 bool vis[maxn], have_failed[maxn]; 10 11 void dfs(int m) { 12 if (m == 1) return; 13 14 memset(vis, false, sizeof(vis)); 15 16 for (int i = 2; i <= n; i++) { 17 if (have_failed[i] || vis[i]) continue; 18 if (gra[1][i] == ‘0‘) { 19 for (int j = 2; j <= n; j++) { 20 if (have_failed[j] || vis[j]) continue; 21 if (gra[1][j] == ‘1‘ && gra[j][i] == ‘1‘) { 22 vis[j] = vis[i] = true; 23 have_failed[i] = true; 24 printf("%d %d\n", i, j); 25 break; 26 } 27 } 28 } 29 } 30 31 for (int i = 2; i <= n; i++) { 32 if (have_failed[i] || vis[i]) continue; 33 if (gra[1][i] == ‘1‘) { 34 vis[i] = true; 35 have_failed[i] = true; 36 printf("%d %d\n", 1, i); 37 break; 38 } 39 } 40 41 int flag = 0, pre = 0; 42 for (int i = 2; i <= n; i++) { 43 if (have_failed[i] || vis[i]) continue; 44 if (gra[1][i] == ‘0‘) { 45 if (!flag) { 46 flag = 1; 47 pre = i; 48 } 49 else { 50 flag = 0; 51 vis[i] = vis[pre] = true; 52 printf("%d %d\n", pre, i); 53 if (gra[pre][i] == ‘0‘) have_failed[pre] = true; 54 else have_failed[i] = true; 55 } 56 } 57 } 58 59 flag = 0; 60 for (int i = 2; i <= n; i++) { 61 if (have_failed[i] || vis[i]) continue; 62 if (!flag) { 63 flag = 1; 64 pre = i; 65 } 66 else { 67 flag = 0; 68 vis[i] = vis[pre] = true; 69 printf("%d %d\n", pre, i); 70 if (gra[pre][i] == ‘0‘) have_failed[pre] = true; 71 else have_failed[i] = true; 72 } 73 } 74 75 dfs(m >> 1); 76 } 77 78 int main() 79 { 80 //freopen("input.txt", "r", stdin); 81 //freopen("output.txt", "w", stdout); 82 while (~scanf("%d", &n)) { 83 memset(have_failed, false, sizeof(have_failed)); 84 for (int i = 1; i <= n; i++) { 85 scanf("%s", gra[i] + 1); 86 } 87 88 dfs(n); 89 } 90 return 0; 91 }
标签：lin void title ems pac git include flag http
原文地址：https://www.cnblogs.com/npugen/p/9678372.html
