CodeForces 1535D : Playoff Tournament 二叉树

传送门

题意

分析

模拟一颗二叉树，找到待修改的节点然后向上维护即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
char str[N];
int tr[N];
int id[N];
int pos[N];
int n,m;void bfs(){queue<int> q;q.push(1);int k = pow(2,n) - 1,p = k;while(q.size()){int t = q.front();q.pop();pos[k] = t;id[t] = k--;int x = t << 1 | 1,y = t << 1;if(x <= p) q.push(x),q.push(y);}
}void build(int u){if(u > pow(2,n) - 1) {tr[u] = 1;return;}build(u << 1),build(u << 1 | 1);if(str[id[u]] == '0'){tr[u] = tr[u << 1];}if(str[id[u]] == '1'){tr[u] = tr[u << 1 | 1];}if(str[id[u]] == '?'){tr[u] = tr[u << 1] + tr[u << 1 | 1];}
}void dfs(int u){if(str[id[u]] == '0'){tr[u] = tr[u << 1];}if(str[id[u]] == '1'){tr[u] = tr[u << 1 | 1];}if(str[id[u]] == '?'){tr[u] = tr[u << 1] + tr[u << 1 | 1];}if(u == 1) return;dfs(u / 2);
}int main() {read(n);scanf("%s",str + 1);read(m);bfs();build(1);while(m--){int id;char a[3];scanf("%d%s",&id,a);str[id] = *a;dfs(pos[id]);di(tr[1]);}return 0;
}

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CodeForces 1535D : Playoff Tournament 二叉树

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题意

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代码

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