ZOJ-1610 + segment tree the two query methods (hematemesis force)

ZOJ-1610 + segment tree the two query methods (hematemesis force)

The meaning of problems

Question is intended: to a n, n represents operations, each operation represents the next segment [l, r] k sections painted colors wherein, l, r, k ranges are 0-8000

This problem is relatively pit can not take n achievements, otherwise it will segmentation fault, I was wrong a lot of times to see the blog know, you must take the 8000 achievements, that is, trees are fixed.

A problem-solving ideas

Segment tree to maintain the color point

A single point of inquiry, a point of a query from left to right, this is a bit violent, but because each query is \ (logn \) is not great, it runs a bit is 60ms

Code to achieve a

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=8e3+7;
struct node{
    int l, r;
    int lazy;
}a[maxn<<2];
int ans[8007];
void down(int rt)
{
    int l=rt<<1, r=rt<<1|1;
    
    a[l].lazy=a[rt].lazy;
    
    a[r].lazy=a[rt].lazy;
    
    a[rt].lazy=-1; 
}
void build(int rt, int l, int r)
{
    a[rt].l=l;
    a[rt].r=r;
    a[rt].lazy=-1; //表示当前点的范围颜色不统一
    if(l==r) return ;
    int mid=(l+r)>>1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid+1, r);
}
void update(int rt, int L, int R, int x)
{
    if(L<=a[rt].l && a[rt].r<=R)
    {
        a[rt].lazy=x;
        return ;
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(a[rt].lazy!=-1)
        down(rt);
    if(L<=mid) update(rt<<1, L, R, x);
    if(R>mid) update(rt<<1|1, L, R, x);
}
int query(int rt, int x) //单点查询
{
    if(a[rt].l==a[rt].r)
    { 
        return a[rt].lazy;
    }
    if(a[rt].lazy!=-1) //记得一定要down一下，因为是单点查询，如果不down，这个点可能还是上一次的状态
        down(rt);
    int mid=(a[rt].l+a[rt].r)>>1;
    if(x<=mid) return query(rt<<1, x);
    else return query(rt<<1|1, x);
}
int main()
{
    int n, a, b, c;
    while(scanf("%d", &n)!=EOF)
    {
        build(1, 1, 8000);
        memset(ans, 0, sizeof(ans));
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            update(1, a+1, b, c);
        }
        int p1, p2;
        p1=query(1, 1);
        if(p1!=-2 && p1!=-1) ans[p1]++;
        for(int i=2; i<=8000; i++)
        {
            p2=query(1, i);
            if(p1!=p2){
                ans[p2]++;
            }
            p1=p2;
        }
        for(int i=0; i<=8000; i++)
        {
            if(ans[i]!=0) 
            {
                printf("%d %d\n", i, ans[i]);
            }
        }
        printf("\n");
    }
    return 0;
}

Two problem-solving ideas

Similarly query interval, if a lazy mark a point of not -1, this interval that we are all one color, they do not down recursion

If the lazy -2, that this region is not a color, an area corresponding to the separated intermediate.

Inquiry is carried out from left to right, pre represents a previous point in the color of the left, it is necessary to determine the color point of the current point and the left are the same.

This code is running 50ms.

Code to achieve two

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=8e3+7;
struct node{
    int l, r;
    int lazy;
}a[maxn<<2];
int ans[8007];
int pre; //pre表示先前块的颜色
void down(int rt)
{
    int l=rt<<1, r=rt<<1|1;
    
    a[l].lazy=a[rt].lazy;
    
    a[r].lazy=a[rt].lazy;
    
    a[rt].lazy=-1; 
}
void build(int rt, int l, int r)
{
    a[rt].l=l;
    a[rt].r=r;
    a[rt].lazy=-2; //表示没有颜色
    if(l==r) return ;
    int mid=(l+r)>>1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid+1, r);
}
void update(int rt, int L, int R, int x)
{
    if(L<=a[rt].l && a[rt].r<=R)
    {
        a[rt].lazy=x;
        return ;
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(a[rt].lazy!=-1 && a[rt].lazy!=-2)
        down(rt);
    if(L<=mid) update(rt<<1, L, R, x);
    if(R>mid) update(rt<<1|1, L, R, x);
    if(a[rt<<1].lazy==a[rt<<1|1].lazy)//相当于up 看下属左右两点的颜色是否相同
    {
        a[rt].lazy=a[rt<<1].lazy; //相同的话，这个点就可以代表下面的两个点
    }
    else a[rt].lazy=-1;//否则-1表示下属两个点的颜色不统一，需要以后特殊标记
}
void query(int rt)//先当于区间查询，从左到右
{
    if(a[rt].lazy==-2) //说明这个点控制的区域没有刷颜色，那么它右边的点也可以和左边的点有相同颜色
    {
        pre=-1;//这个标记-1
        return ;    //结束
    } 
    if(a[rt].lazy!=-1)
    { 
        if(a[rt].lazy!=pre)
        {
            ans[a[rt].lazy]++;
            pre=a[rt].lazy;
        }
        return ;
    }
    query(rt<<1);
    query(rt<<1|1);
}
int main()
{
    int n, a, b, c;
    while(scanf("%d", &n)!=EOF)
    {
        build(1, 1, 8000);
        memset(ans, 0, sizeof(ans));
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            update(1, a+1, b, c);
        }
        pre=-2;
        query(1);
        for(int i=0; i<=8000; i++)
        {
            if(ans[i]!=0) 
            {
                printf("%d %d\n", i, ans[i]);
            }
        }
        printf("\n");
    }
    return 0;
}