Educational Codeforces Round 110 (Rated for Div. 2) D. Playoff Tournament
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题意:给你含有2^k-1个节点的满二叉树,对于叶子节点权值:’?‘ = 2, ‘1’ = 1, ‘0’ = 1; 对于非叶子节点 ’?‘ = lsn + rsn, ‘1’ = rsn, ‘0’ = lsn(lsn, rsn分别对应左右节点的权值)。有q次查询,每次把某个节点的字符修改为 ?1 0 中的一个,求修改后根节点的点权。
思路:因为只修改一个点,那么这个点影响的只有它的父亲节点,所以我们把每个节点的点权全部记录下来,然后对于修改后的节点向上递归,一直到根。 边递归边修改。k 最大为18 所以最多向上递归18次。O(18q);
#include<bits/stdc++.h>
using namespace std;
#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<double, double> PD;
const int MAXN = (1 << 18) + 10;
const int MAX_LEN = 100000 + 10;
const int MAX_LOG_V = 22;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const ull B = 100000007;
int k;
char s[MAXN], w[MAXN];
int no[MAXN], val[MAXN];
void update(int v) {
if( (1<<(k-1)) <= v && v <= (1<<k) - 1) {
if(w[v] == '?') val[v] = 2;
else val[v] = 1;
}
else {
if(w[v] == '?') val[v] = val[v*2]+val[v*2+1];
else if(w[v] == '1') val[v] = val[v*2+1];
else val[v] = val[v*2];
}
}
//向上更新
void dfs(int v) {
update(v);
if(v == 1) return ;
else dfs(v/2);
}
void solve() {
scanf("%d", &k);
scanf("%s", s+1);
int cur = (1<<k) - 1, cnt = 1;
queue<int> que;
que.emplace(1);
//转换为树
while(cur > 0) {
for(int i = cur; i < cur + cnt; i++) {
int v = que.front(); que.pop();
no[i] = v;
w[v] = s[i];
que.emplace(v*2); que.emplace(v*2 + 1);
}
cnt *= 2;
cur -= cnt;
}
for(int i = (1 << k) - 1; i >= 1; i--) {
update(i);
}
int q; scanf("%d", &q);
while(q--) {
int p; char op[5];
scanf("%d %s", &p, op);
w[no[p]] = *op;
update(no[p]);
//如果不为根 就向上更新
if(no[p] != 1) dfs(no[p]/2);
printf("%d\\n", val[1]);
}
}
int main() {
//ios::sync_with_stdio(false);
int t = 1; //scanf("%d", &t);
while(t--) {
solve();
}
return 0;
}
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