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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
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 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 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 |
给你n张画报要贴在墙上，后面贴上去的会覆盖前面贴上去的，求最后能够看见多少张画报。
题目给的区间最大是10000000，所以需要将区间离散化一下，然后线段树求解。
//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