目录
题意:给定一棵树的公司职员管理图,有两种操作,
第一种是 T x y,把 x 及员工都变成 y,
第二种是 C x 询问 x 当前的数。
析:先把该树用dfs遍历,形成一个序列,然后再用线段树进行维护,很简单的线段树。
代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 | #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair< int , int > P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 0x3f3f3f3f3f3f; const double PI = acos (-1.0); const double eps = 1e-8; const int maxn = 5e4 + 10; const int mod = 1000000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = { "0000" , "0001" , "0010" , "0011" , "0100" , "0101" , "0110" , "0111" , "1000" , "1001" , "1010" , "1011" , "1100" , "1101" , "1110" , "1111" }; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in( int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } vector< int > G[maxn]; int in[maxn], out[maxn]; int cnt; int sum[maxn<<2], setv[maxn<<2]; void dfs( int u){ in[u] = ++cnt; for ( int i = 0; i < G[u].size(); ++i) dfs(G[u][i]); out[u] = cnt; } void push_down( int rt){ if (setv[rt] == -1) return ; int l = rt<<1, r = rt<<1|1; sum[l] = sum[r] = setv[rt]; setv[l] = setv[r] = setv[rt]; setv[rt] = -1; } void update( int L, int R, int val, int l, int r, int rt){ if (L <= l && r <= R){ sum[rt] = val; setv[rt] = val; return ; } push_down(rt); int m = l + r >> 1; if (L <= m) update(L, R, val, lson); if (R > m) update(L, R, val, rson); } int query( int M, int l, int r, int rt){ if (l == r) return sum[rt]; push_down(rt); int m = l + r >> 1; return M <= m ? query(M, lson) : query(M, rson); } int main(){ int T; cin >> T; for ( int kase = 1; kase <= T; ++kase){ scanf ( "%d" , &n); for ( int i = 1; i <= n; ++i) G[i].clear(); memset (in, 0, sizeof in); for ( int i = 1; i < n; ++i){ int u, v; scanf ( "%d %d" , &u, &v); G[v].push_back(u); ++in[u]; } cnt = 0; for ( int i = 1; i <= n; ++i) if (!in[i]){ dfs(i); break ; } memset (setv, -1, sizeof setv); memset (sum, -1, sizeof sum); scanf ( "%d" , &m); char s[5]; printf ( "Case #%d:\n" , kase); while (m--){ int x, y; scanf ( "%s %d" , s, &x); if (s[0] == 'C' ) printf ( "%d\n" , query(in[x], 1, n, 1)); else { scanf ( "%d" , &y); update(in[x], out[x], y, 1, n, 1); } } } return 0; } |
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