zoukankan      html  css  js  c++  java
  • hdu1199(离散化线段树)

    Color the Ball

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6689    Accepted Submission(s): 1655

    Problem Description
    There are infinite balls in a line (numbered 1 2 3 ....), and initially all of them are paint black. Now Jim use a brush paint the balls, every time give two integers a b and follow by a char 'w' or 'b', 'w' denotes the ball from a to b are painted white, 'b' denotes that be painted black. You are ask to find the longest white ball sequence.
     
    Input
    First line is an integer N (<=2000), the times Jim paint, next N line contain a b c, c can be 'w' and 'b'.

    There are multiple cases, process to the end of file.
     
    Output
    Two integers the left end of the longest white ball sequence and the right end of longest white ball sequence (If more than one output the small number one). All the input are less than 2^31-1. If no such sequence exists, output "Oh, my god".
     
    Sample Input
    3 1 4 w 8 11 w 3 5 b
     
    Sample Output
    8 11
     
     
     
    题目大意:区间初始化颜色为黑色,给定N个区间颜色,求白色最长区间的L,R。
     
     
    AC代码:
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    
    const int maxn=2e5+5;
    int cnt=0;
    
    int l[maxn*2],node[maxn*4];
    struct node
    {
        int l,r,color,lazy;//color=1白色 
    }init[maxn],tree[4*maxn],d[maxn];
    
    void reset_node()
    {
        int i=1,j=1;
        node[1]=l[1];
        for(i=2;i<=cnt;i++)
        {
            if(node[j]==l[i]) continue;
            if(node[j]+1<l[i])
            {
                node[++j]=node[j-1]+1;
            }
            if(node[j]<l[i]-1)
            {
                node[++j]=l[i]-1;
            }
            node[++j]=l[i];
        }
        cnt=j;
    }
    
    void btree(int l,int r,int k)
    {
        tree[k].color=0;
        tree[k].lazy=1;
        tree[k].l=l;tree[k].r=r;
        if(l==r) return ;
        int mid=(l+r)/2;
        btree(l,mid,k<<1);
        btree(mid+1,r,k<<1|1);
    }
    
    int ccnt=0;
    void date_tree(int l,int r,int color,int k)
    {
        //printf("%d ",ccnt++);
        int mid=(tree[k].l+tree[k].r)/2;
        if(tree[k].l>=l&&tree[k].r<=r)
        {
            tree[k].color=color;
            tree[k].lazy=1;
            return ;
        }
        if(tree[k].lazy&&tree[k].color==color)
            return ;
        if(tree[k].lazy)
        {
            tree[k<<1].color=tree[k<<1|1].color=tree[k].color;
            tree[k<<1].lazy=tree[k<<1|1].lazy=1;
        }
        if(l>mid)
        {
            date_tree(l,r,color,k<<1|1);
        }
        else if(r<=mid)
        {
            date_tree(l,r,color,k<<1);
        }
        else
        {
            date_tree(l,r,color,k<<1);
            date_tree(l,r,color,k<<1|1);
        }
        tree[k].lazy=0;
    }
    
    int in_node(int k)
    {
        int l=1,r=cnt,mid;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(k==node[mid]) break;
            if(k>node[mid]) l=mid+1;
            if(k<node[mid]) r=mid-1;
        }
        return mid;
    }
    
    void push_down(int k,int l,int r)
    {
        int mid=(l+r)/2;
        if(tree[k].l==tree[k].r) return;
        if(tree[k].lazy)
        {
            tree[k<<1].color=tree[k<<1|1].color=tree[k].color;
            tree[k<<1].lazy=tree[k<<1|1].lazy=1;
        }
        push_down(k<<1,tree[k<<1].l,tree[k<<1].r);
        push_down(k<<1|1,tree[k<<1|1].l,tree[k<<1|1].r);
    }
    
    int dn=0;
    void get_node(int l,int r,int k)
    {
        if(tree[k].lazy)
            d[++dn]=tree[k];
        if(l==r) return ;
        int mid=(l+r)/2;
        get_node(l,mid,k<<1);
        get_node(mid+1,r,k<<1|1);
    }
    
    int main()
    {
        int n;
        cnt=0;
        char s[5];
        while(scanf("%d",&n)!=EOF)
        {
            //printf("1
    ");
            cnt=0;
            for(int i=1;i<=n;i++)
            {
                scanf("%d%d%s",&init[i].l,&init[i].r,s);
                l[++cnt]=init[i].l;l[++cnt]=init[i].r;
                if(s[0]=='w')
                {
                    init[i].color=1;
                }
                else init[i].color=0;
            }
            sort(l+1,l+1+cnt);
            reset_node();
            //printf("%d...
    ",cnt);
            btree(1,cnt,1);
            //printf("2
    ");
            int ll,rr,cc;
            for(int i=1;i<=n;i++)
            {
                ll=in_node(init[i].l);
                rr=in_node(init[i].r);
                cc=init[i].color;
                date_tree(ll,rr,cc,1);
            }
            //printf("3
    ");
            push_down(1,1,cnt);
            //printf("4
    ");
            dn=0;
            get_node(1,cnt,1);
            int L,R;
            int len=-1;
            //printf("5
    ");
            for(int i=1;i<=dn;i++)
            {
                if(d[i].color)
                {
                    int j=i;
                    while(d[i+1].color!=0&&i+1<=dn) i++;
                    if(i==dn&&node[d[i].r]-node[d[j].l]+1>len)
                    {
                        L=node[d[j].l];
                        R=node[d[i].r];
                        len=R-L+1;
                    }
                    if(i<dn&&node[d[i].l]-node[d[j].l]>len)
                    {
                        L=node[d[j].l];
                        R=node[d[i].l];
                        len=R-L;
                    }
                }
            }
            //for(int i=1;i<=11;i++)
            //printf("%d
    ",tree[i].color);
            //printf("6
    ");
            if(len!=-1)
                printf("%d %d
    ",L,R);
            else
                printf("Oh, my god
    ");
        }
        return 0;
    }
    View Code
  • 相关阅读:
    web应用后台开发的故事
    XML的定义、用途、以及它的发展前景和存在的问题等等
    本学期(大三下学期)学习目标
    企业级应用与互联网应用的区别?
    新能源汽车无线充电管理网站4
    新能源汽车无线充电管理网站3
    新能源汽车无线充电管理网站2
    企业级应用与互联网应用的区别
    javaee 新学期新目标
    团队项目PCP--自我评价
  • 原文地址:https://www.cnblogs.com/noback-go/p/11239981.html
Copyright © 2011-2022 走看看