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  • 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
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  • 原文地址:https://www.cnblogs.com/noback-go/p/11239981.html
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