codeforces 558D Guess Your Way Out! II 规律

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题目链接

题意：

给出n和q

 像这样标号

首先把所有提示的区间都映射到叶子节点上

先把一定不在的问题转成2个一定存在的提示。

那么显然每个提示里都包含了出口，所以我们查询一下哪个点是被q个区间覆盖了，则这个点就是出口。

#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <cstdio>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<ll, ll> pii;
const int N = 500005;
const int inf = 1e9 + 10;
int n, q;

ll L[55], R[55];
map<ll, int>mp;
int main() {
	L[1] = R[1] = 1;
	for (int i = 2; i <= 50; i++)L[i] = L[i - 1] << 1, R[i] = R[i - 1] << 1 | 1;
	rd(n); rd(q);
	if (q == 0) {
		if (n == 1)puts("1");
		else puts("Data not sufficient!");
		return 0;
	}
	for (int i = 0, dep, ok; i < q; i++) {
		ll l, r;
		rd(dep); rd(l); rd(r); rd(ok);
		while (dep < n) {
			l <<= 1;
			r = r << 1 | 1;
			dep++;
		}
		if (ok)mp[l]++, mp[r + 1]--;
		else {
			mp[L[n]]++; mp[l]--;
			mp[r + 1]++; mp[R[n]+1]--;
		}
	}	
	int sum = 0;
	ll pre = -1, cnt = 0, ans = 0;
	for (auto i : mp) {
		sum += i.second;
		if (pre != -1) {
			cnt += i.first - pre;
			ans = pre;
		}
		if (sum == q)pre = i.first;
		else pre = -1;
	}
	if (cnt == 0)puts("Game cheated!");
	else if (cnt > 1)puts("Data not sufficient!");
	else pt(ans);
	return 0;
}

codeforces 558D Guess Your Way Out! II 规律

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原文地址：http://blog.csdn.net/qq574857122/article/details/46896861