CSU 1784 Internet of Lights and Switches 位运算+思维

最新推荐文章于 2018-07-12 10:07:00 发布

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最新推荐文章于 2018-07-12 10:07:00 发布

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分类专栏： ACM-思维文章标签： codeforces acm 位运算

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1784: Internet of Lights and Switches

Time Limit: 5 Sec Memory Limit: 128 Mb Submitted: 246 Solved: 37

Description

You are a fan of "Internet of Things"(IoT, 物联网), so you build a nice Internet of Lights and Switches in your huge mansion. Formally, there are n lights and m switches, each switch controls one or more lights, i.e. pressing that switch flips the status of those lights (on->off, off->on).

Initially, all the lights are on. Your task is to count the number of ways to turn off all the lights by pressing someconsecutive switches. Each switch should not be pressed more than once. There is only one restriction: the number of switches you pressed should be between a and b (inclusive).

Input

There will be at most 20 test cases. Each test case begins with a line containing four integers n, m, a, b (2<=n<=50, 1<=a<=b<=m<=300000). Each of the following m lines contains a 01 string of length n. The i-th character is 1 if and only if that switch controls the i-th light. The size of the whole input file does not exceed 8MB.

Output

For each test case, print the case number, and the number of ways to turn off all the lights.

Sample Input

Sample Output

Case 1: 3
Case 2: 0
Case 3: 2

让l,r区间能把当前区间所有等的状态调整到已关状态，那么一定是1,l和1,r两个区间把最初的状态调整到了全关的状态。所以求前缀异或和。枚举每一个合法区间的终点，将其所在前缀异或和的状态抑或初状态，可得其起点应有的状态。如果我们在一开始计算前缀和的时候用哈希记录每种状态对应的开关编号，那么这在这一步可以使用二分搜索计算得出合法的起点数，也就是答案数的一部分。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<map>
#include<cmath>
#include<vector>
using namespace std;

long long n,m,a,b,pre[300005];
map<long long ,vector<int> > mapp;

int main()
{
    int cas=0;
    while (~scanf("%d%d%d%d",&n,&m,&a,&b))
    {
        cas++;
        long long st=((long long)1<<n)-1;
        mapp.clear();
        mapp[st].push_back(0);
        pre[0]=st;
        for (int i=1; i<=m; i++)
        {
            long long cnt=0;
            for (int j=1; j<=n; j++)
            {
                char ch; cin>>ch;
                if (ch=='1')
                {
                    cnt+=((long long)1<<(j-1));
                }
            }
            pre[i]=pre[i-1]^cnt;
            mapp[pre[i]].push_back(i);
        }
        int ans=0;
        for (int i=a; i<=m; i++)
        {
            long long tmp=st^pre[i];
            if (mapp[tmp].size()==0) continue;
            int pos1=lower_bound(mapp[tmp].begin(),mapp[tmp].end(),i-b)-mapp[tmp].begin();
            int pos2=upper_bound(mapp[tmp].begin(),mapp[tmp].end(),i-a)-mapp[tmp].begin();
            ans+=abs(pos1-pos2);
        }
        cout<<"Case "<<cas<<": "<<ans<<endl;
    }
    return 0;
}