标签：href ems 输出 view stdio.h 技术分享 clu scan ble
题目链接: http://codeforces.com/problemset/problem/780/E
题意: 给出一个 n 个点 m 条边的图, 有 k 个人, 初始位置可以为任意位置, 每个人最多不能经过超过 ceil(2 * n / k) 个顶点. 要使 k 个人经历所有顶点, 并输出 k 个人经历的顶点数目和路径.
思路: 先 dfs 遍历一下图, 并记录路径 (回溯路径也要记录). 然后再将路径按条件分配给 k 个人即可.
代码:
1 #include <iostream> 2 #include <stdio.h> 3 #include <vector> 4 using namespace std; 5 6 const int MAXN = 2e5 + 10; 7 vector<int> vt[MAXN]; 8 int vis[MAXN], path[MAXN << 2]; 9 int indx = 0; 10 11 void dfs(int x){ 12 vis[x] = 1; 13 path[++indx] = x; 14 for(int i = 0; i < vt[x].size(); i++){ 15 if(!vis[vt[x][i]]){ 16 dfs(vt[x][i]); 17 path[++indx] = x; 18 } 19 } 20 } 21 22 int main(void){ 23 int n, m, k, x, y; 24 scanf("%d%d%d", &n, &m, &k); 25 for(int i = 0; i < m; i++){ 26 scanf("%d%d", &x, &y); 27 vt[x].push_back(y); 28 vt[y].push_back(x); 29 30 } 31 dfs(1); 32 int cnt = (2 * n + k - 1) / k; 33 for(int i = 0; i < k; i++){ 34 int cc = min(cnt, indx); 35 if(!cc){ 36 printf("1 1\n"); 37 continue; 38 } 39 printf("%d ", cc); 40 for(int j = 0; j < cc && indx; j++,indx--){ 41 printf("%d ", path[indx]); 42 } 43 puts(""); 44 } 45 return 0; 46 }
标签：href ems 输出 view stdio.h 技术分享 clu scan ble
原文地址：http://www.cnblogs.com/geloutingyu/p/7162035.html
