标签：
|
Mayor‘s posters
Description
The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters
and introduce the following rules:
They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. Your task is to find the number of visible posters when all the posters are placed given the information about posters‘ size, their place and order of placement on the electoral wall. Input
The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among
the n lines contains two integer numbers li and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= li <= ri <= 10000000. After
the i-th poster is placed, it entirely covers all wall segments numbered li, li+1 ,... , ri.
Output
For each input data set print the number of visible posters after all the posters are placed.
The picture below illustrates the case of the sample input.  Sample Input 1 5 1 4 2 6 8 10 3 4 7 10 Sample Output 4 Source |
//35616K 94MS
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 10007
using namespace std;
int id[M*2],_hash[10000007];
int n,num;
struct T
{
int left,right,mid,color;
T(){color=0;}
}tree[M<<4];
struct P
{
int l,r;
}p[M*8];
void build(int l,int r,int i)
{
tree[i].left=l;tree[i].right=r;
tree[i].mid=(l+r)>>1;tree[i].color=0;
if(l==r)return;
build(l,tree[i].mid,i<<1);
build(tree[i].mid+1,r,i<<1|1);
}
bool query(int l,int r,int i)
{
bool flag;
if(tree[i].color)return false;
if(tree[i].left==l&&tree[i].right==r)
{
tree[i].color=true;
return true;
}
if(r<=tree[i].mid)flag=query(l,r,i<<1);
else if(l>tree[i].mid)flag=query(l,r,i<<1|1);
else
{
bool flag1=query(l,tree[i].mid,i<<1);
bool flag2=query(tree[i].mid+1,r,i<<1|1);
flag=flag1||flag2;
}
if(tree[i<<1].color&&tree[i<<1|1].color)//反馈原节点
tree[i].color=true;
return flag;
}
void lisanhua()
{
sort(id,id+num);
num=unique(id,id+num)-id;
for(int i=0;i<num;i++)
_hash[id[i]]=i;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
num=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&p[i].l,&p[i].r);
id[num++]=p[i].l;
id[num++]=p[i].r;
}
lisanhua();
build(0,num-1,1);
int count=0;
for(int i=n;i>=1;i--)//从最后贴的往前找
{
if(query(_hash[p[i].l],_hash[p[i].r],1))
count++;
}
printf("%d\n",count);
}
return 0;
}
POJ 2528 Mayor's posters 区间离散化线段树
标签：
原文地址：http://blog.csdn.net/crescent__moon/article/details/43604787
