[HDOJ3974]Assign the task（建树胡搞）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3974

出现在窝bin的线段树专题里…第一时间想的是记录入度找出根节点，然后标记深度转换到线段树中。但是发现并不会这个姿势…于是直接深搜更新了…发现没有退化成链的数据。于是水水地过去了………………

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <iomanip>
  4 #include <cstring>
  5 #include <climits>
  6 #include <complex>
  7 #include <fstream>
  8 #include <cassert>
  9 #include <cstdio>
 10 #include <bitset>
 11 #include <vector>
 12 #include <deque>
 13 #include <queue>
 14 #include <stack>
 15 #include <ctime>
 16 #include <set>
 17 #include <map>
 18 #include <cmath>
 19 
 20 using namespace std;
 21 
 22 #define fr first
 23 #define sc second
 24 #define pb(a) push_back(a)
 25 #define Rint(a) scanf("%d", &a)
 26 #define Rll(a) scanf("%I64d", &a)
 27 #define Rs(a) scanf("%s", a)
 28 #define FRead() freopen("in", "r", stdin)
 29 #define FWrite() freopen("out", "w", stdout)
 30 #define Rep(i, len) for(int i = 0; i < (len); i++)
 31 #define For(i, a, len) for(int i = (a); i < (len); i++)
 32 #define Cls(a) memset((a), 0, sizeof(a))
 33 #define Clr(a, x) memset((a), (x), sizeof(a))
 34 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 35 
 36 #define lrt rt << 1
 37 #define rrt rt << 1 | 1
 38 const int maxn = 50010;
 39 int n, m;
 40 int in[maxn];
 41 int depth[maxn];
 42 bool vis[maxn];
 43 vector<int> G[maxn];
 44 int q[maxn], front, tail;
 45 int task[maxn];
 46 
 47 void bfs(int rt) {
 48     front = tail = 0;
 49     int lv = 0;
 50     depth[rt] = lv++; q[tail++] = rt; vis[rt] = 1;
 51     while(front < tail) {
 52         int u = q[front++];
 53         Rep(i, G[u].size()) {
 54             if(!vis[G[u][i]]) {
 55                 vis[G[u][i]] = 1;
 56                 depth[G[u][i]] = lv;
 57                 q[tail++] = G[u][i];
 58             }
 59         }
 60         lv++;
 61     }
 62 }
 63 
 64 void update(int rt, int p) {
 65     task[rt] = p;
 66     Rep(i, G[rt].size()) {
 67         update(G[rt][i], p);
 68     }
 69 }
 70 
 71 int main() {
 72     // FRead();
 73     int T, _ = 1;
 74     Rint(T);
 75     int u, v;
 76     char cmd[6];
 77     while(T--) {
 78         Rint(n);
 79         Rep(i, n+2) G[i].clear();
 80         Cls(depth); Cls(vis); Cls(in); Clr(task, -1);
 81         Rep(i, n-1) {
 82             Rint(u); Rint(v);
 83             G[v].push_back(u); in[u]++;
 84         }
 85         int rt = -1;
 86         For(i, 1, n+1) if(!in[i]) rt = i;
 87         bfs(rt);
 88         Rint(m);
 89         printf("Case #%d:\n", _++);
 90         while(m--) {
 91             Rs(cmd);
 92             if(cmd[0] == 'C') {
 93                 Rint(u);
 94                 printf("%d\n", task[u]);
 95             }
 96             if(cmd[0] == 'T') {
 97                 Rint(u); Rint(v);
 98                 update(u, v);
 99             }
100         }
101     }
102     return 0;
103 }

 

posted @ 2016-05-16 14:27  Kirai  阅读(173)  评论(0)    收藏  举报