Internet of Lights and Switches(MAP记录+二分) 2015年湖南省赛第 I 题

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 some consecutive 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).

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.

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

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

题意：有n个灯最初是亮的，现在要通过连续的操作使得灯全灭，现在m个操作,可以连续的操作个数为a~b，问有多少种情况使得灯全灭。

解题：用map<long long , vector<int> >记录每种灯的状态所对应第 i 个操作（第i个操作的状态为：输入的前i个操作的异或状态） + 二分.

#include<stdio.h>
#include<string.h>
#include<string>
#include<map>
#include<vector>
using namespace std;
const int N = 300010 ;
#define ll long long
ll num[N] ;
map<ll, vector<int> >mp;
void two(ll tnum , int n , int& L , int& R )
{int l=0,r=mp[tnum].size()-1 , mid;while(l<=r){mid = (l+r)>>1;if(mp[tnum][mid]>=L)r = mid - 1 ;elsel = mid + 1;}L  = l ;if(L==mp[tnum].size()){L = -1 ;R =  L - 1 ;return ;}l = 0 , r = mp[tnum].size()-1 ;while(l<=r){mid = (l+r)>>1;if(mp[tnum][mid]>=R)r = mid - 1 ;elsel = mid + 1;}if(l==mp[tnum].size())R = l-1;else if(mp[tnum][l]>R)R = l-1 ;elseR = l ;
}int main()
{int n,m,a,b , ans , T = 0;char s[100] ;while(scanf("%d%d%d%d",&n,&m,&a,&b)>0){mp[0].push_back(0);num[0]=0;for(int i=1; i<=m; i++){scanf("%s",s);num[i] = 0 ;for(int j=0; j<n; j++)if(s[j]=='1'&& !(num[i-1]&(1LL<<j)) || s[j]=='0'&&(num[i-1]&(1LL<<j))==(1LL<<j))num[i] |= 1LL<<j ;mp[ num[i] ].push_back(i) ;// printf("size = %d\n",mp[num[i]].size());}ans = 0 ;for(int i=1; i<=m; i++){ll tnum = 0 ;for(int j=0; j<n; j++)if((num[i]&(1LL<<j))==0)tnum |= 1LL<<j;int L = i-b , R = i-a ;two(tnum , n , L , R );if(R>=L)ans += R-L+1 ;// printf("[ %d , %d ] num = %lld ,tnum = %lld size = %d\n",L,R,num[i],tnum,mp[tnum].size());}for(int i=0; i<=m; i++)mp[num[i]].clear() ;printf("Case %d: %d\n",++T , ans ) ;}
}