P2846 [USACO08NOV]光开关Light Switching

题目描述

Farmer John tries to keep the cows sharp by letting them play with intellectual toys. One of the larger toys is the lights in the barn. Each of the N (2 <= N <= 100,000) cow stalls conveniently numbered 1..N has a colorful light above it.

At the beginning of the evening, all the lights are off. The cows control the lights with a set of N pushbutton switches that toggle the lights; pushing switch i changes the state of light i from off to on or from on to off.

The cows read and execute a list of M (1 <= M <= 100,000) operations expressed as one of two integers (0 <= operation <= 1).

The first kind of operation (denoted by a 0 command) includes two subsequent integers S_i and E_i (1 <= S_i <= E_i <= N) that indicate a starting switch and ending switch. They execute the operation by pushing each pushbutton from S_i through E_i inclusive exactly once.

The second kind of operation (denoted by a 1 command) asks the cows to count how many lights are on in the range given by two integers S_i and E_i (1 <= S_i <= E_i <= N) which specify the inclusive range in which the cows should count the number of lights that are on.

Help FJ ensure the cows are getting the correct answer by processing the list and producing the proper counts.

灯是由高科技——外星人鼠标操控的。你只要左击两个灯所连的鼠标，

这两个灯，以及之间的灯都会由暗变亮，或由亮变暗。右击两个灯所连的鼠

标，你就可以知道这两个灯，以及之间的灯有多少灯是亮的。起初所有灯都是暗的，你的任务是在LZ之前算出灯的亮灭。

输入输出格式

输入格式：

第1 行: 用空格隔开的两个整数N 和M，n 是灯数

第2..M+1 行: 每行表示一个操作, 有三个用空格分开的整数: 指令号, S_i 和E_i

第1 种指令(用0 表示)包含两个数字S_i 和E_i (1 <= S_i <= E_i <= N), 它们表示起

始开关和终止开关. 表示左击

第2 种指令(用1 表示)同样包含两个数字S_i 和E_i (1 <= S_i <= E_i <= N), 不过这

种指令是询问从S_i 到E_i 之间的灯有多少是亮着的.

输出格式：

输入输出样例

输入样例#1： 

4 5
0 1 2
0 2 4
1 2 3
0 2 4
1 1 4

输出样例#1： 

1
2

说明

原题时间限制为2s，内存限制为16M

Solution：

　　本题比较水（是真的很水）。

　 和以前做过的某道好像叫做开关的题目一模一样。

   线段树维护区间和，区间修改时等价于区间取反，$sum$变为长度减去原来的和，打个异或懒惰标记（两次操作等于没有操作）。

　 练手瞎搞一弃就好了。

代码：

#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define il inline
using namespace std;

const int N=100005;
int n,m,sum[N*3],lazy[N*3];

il int gi(){
    int a=0;char x=getchar();
    while(x<'0'||x>'9')x=getchar();
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return a;
}

il void pushup(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}

il void pushdown(int rt,int len){
    if(lazy[rt]){
        lazy[rt<<1]^=1;lazy[rt<<1|1]^=1;
        sum[rt<<1]=(len-(len>>1))-sum[rt<<1];
        sum[rt<<1|1]=(len>>1)-sum[rt<<1|1];
        lazy[rt]=0;
    }
}

il void update(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r){sum[rt]=(r-l+1)-sum[rt];lazy[rt]^=1;return;}
    pushdown(rt,r-l+1);
    int m=l+r>>1;
    if(L<=m)update(L,R,lson);
    if(R>m)update(L,R,rson);
    pushup(rt);
}

il int query(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r)return sum[rt];
    pushdown(rt,r-l+1);
    int m=l+r>>1,ret=0;
    if(L<=m)ret+=query(L,R,lson);
    if(R>m)ret+=query(L,R,rson);
    return ret;
}

int main(){
    n=gi(),m=gi();
    int f,x,y;
    while(m--){
        f=gi(),x=gi(),y=gi();
        if(!f)update(x,y,1,n,1);
        else printf("%d
",query(x,y,1,n,1));
    }
    return 0;
}