P3870 [TJOI2009]开关

链接:Miku

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凡天下之事，但知其一不知其二者多矣，可据理臆断欤？

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我就是但知其一呜呜呜

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%%%并感谢yyq大佬

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很显然可以用线段树操作

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这次的lazy指的是这个区间需不需要被反转，然后显然，一个区间反转后亮的灯泡的值就是区间灯泡总量-原来亮的值

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所以写出来了个这个

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m;
long long sum[400005], lazy[400005];
int f,x,y;
long long k;
void pushdown(int x, int L, int R){
    if (lazy[x] != 0){
        int mid = (L + R) >> 1;
        lazy[x << 1] =1;
        lazy[x << 1 | 1] = 1;
        sum[x << 1] =  (mid - L + 1)-sum[x<<1];
        sum[x << 1 | 1] =  (R - mid)-sum[x<<1|1];
        lazy[x] = 0;
    }
    return;
}
void pushup(int x){
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
    return;
}
void update(int x, int l, int r, int L, int R){
    if (L <= l && r <= R){
        lazy[x] = 1;
        sum[x] = r - l + 1 - sum[x];
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    if (L <= mid) update(x << 1, l, mid, L, R);
    if (R > mid) update(x << 1 | 1, mid + 1, r, L, R);
    pushup(x);
}
long long query(int x, int l, int r, int L, int R){
    if (L <= l && r <= R){
        return sum[x];
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    long long ans = 0;
    if (L <= mid) ans += query(x << 1, l, mid, L, R);
    if (R > mid) ans += query(x << 1 | 1, mid + 1, r, L, R);
    return ans;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i){
        scanf("%d",&f);
        if(f==0){
            scanf("%d%d",&x,&y);
            update(1, 1, n, x, y);
        }
        else{
            scanf("%d%d",&x,&y);
            printf("%lld\n", query(1, 1, n, x, y));
        }
    }
    return 0;
}

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它过样例了，提交，10分

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？？？？？？？？？？？？？？

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问题出在lazy上

但是如果一个区间被反转了偶数次，其实就相当于没动过，所以我们在处理懒标记的时候需要特判

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#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m;
long long sum[400005], lazy[400005];
int f,x,y;
long long k;
void pushdown(int x, int L, int R){
    if (lazy[x] != 0){
        int mid = (L + R) >> 1;
        lazy[x << 1] =(lazy[x<<1]+1)%2;
        lazy[x << 1 | 1] = (lazy[x<<1|1]+1)%2;
        sum[x << 1] =  (mid - L + 1)-sum[x<<1];
        sum[x << 1 | 1] =  (R - mid)-sum[x<<1|1];
        lazy[x] = 0;
    }
    return;
}
void pushup(int x){
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
    return;
}
void update(int x, int l, int r, int L, int R){
    if (L <= l && r <= R){
        lazy[x] = (lazy[x]+1)%2;
        sum[x] = r - l + 1 - sum[x];
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    if (L <= mid) update(x << 1, l, mid, L, R);
    if (R > mid) update(x << 1 | 1, mid + 1, r, L, R);
    pushup(x);
}
long long query(int x, int l, int r, int L, int R){
    if (L <= l && r <= R){
        return sum[x];
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    long long ans = 0;
    if (L <= mid) ans += query(x << 1, l, mid, L, R);
    if (R > mid) ans += query(x << 1 | 1, mid + 1, r, L, R);
    return ans;
}
int main(){
    scanf("%d% d",&n,&m);
    for(int i=1;i<=m;++i){
        scanf("%d",&f);
        if(f==0){
            scanf("%d%d",&x,&y);
            update(1, 1, n, x, y);
        }
        else{
            scanf("%d%d",&x,&y);
            printf("%lld\n", query(1, 1, n, x, y));
        }
    }
    return 0;
}