BZOJ4245 [ONTAK2015]OR-XOR 【贪心】

题目链接

BZOJ4245

题解

套路①
位运算当然要分位讨论,高位优先
考虑在\(or\)下,如果该位为\(0\),则每一位都为\(0\)

套路②
我们选m段异或和,转化为\(m\)个前缀和的点,且其中有一个是\(n\)

容易发现,该位结果要为0,则选取的前缀和该位都为\(0\)
所以贪心查找所有该位为\(0\)的,首先第\(n\)个前缀和一定要为\(0\),如果其它满足有至少\(m - 1\)个,那么该位答案为\(0\),剩余为\(1\)的打上标记不能选
如果不够,那这一位就没办法了,直接放弃

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int n,m,tag[maxn];
LL a[maxn],ans;
int main(){
	n = read(); m = read();
	REP(i,n) a[i] = read() ^ a[i - 1];
	for (LL i = 1ll << 61; i; i >>= 1){
		if (a[n] & i){
			ans += i;
			continue;
		}
		int cnt = 1;
		for (int j = 1; j < n; j++)
			if (!tag[j] && !(a[j] & i)) cnt++;
		if (cnt >= m){
			for (int j = 1; j < n; j++)
				if (!tag[j] && (a[j] & i))
					tag[j] = true;
		}
		else ans += i;
	}
	printf("%lld\n",ans);
	return 0;
}

posted @ 2018-05-26 19:30  Mychael  阅读(159)  评论(0编辑  收藏  举报