HDU Paint The Wall

Paint The Wall

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 120 Accepted Submission(s): 22

Problem Description

As a amateur artist, Xenocide loves painting the wall. The wall can be considered as a line consisting of n nodes. Each node has its own color.

Xenocide spends all day in front of the wall. Sometimes, he paints some consecutive nodes so that these nodes have the same color. When he feels tired, he focuses on a particular color and counts the number of nodes that have this color within a given interval.

Now Xenocide is tired of counting, so he turns to you for help.

Input

The input consists of several test cases.
The first line of each test case contains two integer n, m(1<=n, m<=100000) indicating the length of the wall and the number of queries.
The following line contains N integers which describe the original color of every position.
Then m lines follow. Each line contains 4 non-negative integers a, l, r, z(1<=a<=2, 0<=l<=r<n ,0<=z<2³¹).
a = 1 indicates that Xenocide paints nodes between l and r and the resulting color is z.
a = 2 indicates that Xenocide wants to know how many nodes between l and r have the color z.

Output

Print the corresponding answer for each queries.

SampleInput

SampleOutput

成段更新,询问过程如果遇到有颜色的线段就可以返回了,修改过程则是遇到有颜色的线段如果匹配就更改,否则就向下传递.但是这么做会TLE.优化的方法是为每条线段增加两个域,记录区间上面颜色的最大值和最小值.如果查询的颜色不在这个范围内,就直接返回. 但是为什么加了一个简单的优化就可以从10000+MS减少到484MS,是否说明测试数据很极端?

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

const int maxn=100010;

struct Tree{
    int l,r;    //左端点、右端点 
    int c;      //color 
    int mini,most;
};

struct Tree a[maxn<<2];
int v[maxn];

void build(int l,int r,int rt){
    int mid=(l+r)>>1;
    a[rt].l=l;
    a[rt].r=r;
    if(l==r){
        a[rt].c=v[l];
        a[rt].mini=v[l];
        a[rt].most=v[l];
        return ;
    }
    build(lson);
    build(rson);
    if(a[rt<<1].c==a[rt<<1|1].c)
        a[rt].c=a[rt<<1].c;
    else
        a[rt].c=-1;
    a[rt].mini=min(a[rt<<1].mini,a[rt<<1|1].mini);
    a[rt].most=max(a[rt<<1].most,a[rt<<1|1].most);
}

void Update(int l,int r,int id,int val){
    if(l==a[id].l && r==a[id].r){
        a[id].c=val;
        a[id].mini=val;
        a[id].most=val;
        return ;
    }
    if(a[id].l==a[id].r)
        return ;
    if(a[id].c!=-1){
        a[id<<1].c=a[id<<1|1].c=a[id].c;
        a[id].c=-1;
        a[id<<1].mini=a[id<<1|1].mini=a[id].mini;
        a[id<<1].most=a[id<<1|1].most=a[id].most;
    }
    int mid=(a[id].l+a[id].r)>>1;
    if(mid>=r)
        Update(l,r,id<<1,val);
    else if(mid<l)
        Update(l,r,id<<1|1,val);
    else{
        Update(l,mid,id<<1,val);
        Update(mid+1,r,id<<1|1,val);
    }
    if(a[id<<1].c==a[id<<1|1].c)
        a[id].c=a[id<<1].c;
    a[id].mini=min(a[id<<1].mini,a[id<<1|1].mini);
    a[id].most=max(a[id<<1].most,a[id<<1|1].most);
}

int ans;

void query(int l,int r,int id,int val){
    if(a[id].c!=-1){
        if(a[id].c==val)
            ans+=r-l+1;
        return ;
    }
    if(a[id].l==a[id].r)
        return ;
    if(a[id].mini>val  || a[id].most<val)    //重要的优化 
        return ;
    int mid=(a[id].l+a[id].r)>>1;
    if(mid>=r)
        query(l,r,id<<1,val);
    else if(mid<l)
        query(l,r,id<<1|1,val);
    else{
        query(l,mid,id<<1,val);
        query(mid+1,r,id<<1|1,val);
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    int n,m;
    while(~scanf("%d%d",&n,&m)){
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++)
            scanf("%d",&v[i]);
        build(0,n-1,1);
        int flag,lt,rt,val;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d%d",&flag,&lt,&rt,&val);
            if(flag==1)
                Update(lt,rt,1,val);
            else{
                ans=0;
                query(lt,rt,1,val);
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}