poj 2528 (线段树+特殊离散化)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 51098 | Accepted: 14788 |
Description
- Every candidate can place exactly one poster on the wall.
- All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
- The wall is divided into segments and the width of each segment is one byte.
- Each poster must completely cover a contiguous number of wall segments.
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
Output
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 思路:
很经典的题。。线段树+离散化,难点主要在于需要用特殊的离散化方法,一般的离散化方法会出错
比如下面这串数据
3
1-10
1-4
6-10
一般离散化后就会变成 a[0] = 1;a[1] = 4;a[2] = 6;a[3]= 10;
离散化将把这些数据的下标作为值放进线段树处理后就会成为:
0 - 3 为1颜色
0 - 1 为2颜色
2 - 3 为3颜色
最后只存在两种颜色
但正确的过程应该是:
1 - 10 为1颜色
1 - 4 为2颜色
6 - 10 为3颜色
最后存在三种颜色 实现代码:
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 2e5+;
int col[M<<],Hash[M<<],cnt,a[M],l[M],r[M];
void pushdown(int rt){
if(col[rt]!=-){
col[rt<<] = col[rt<<|] = col[rt];
col[rt] = -;
}
} void update(int L,int R,int c,int l,int r,int rt){
if(L <= l&&R >= r){
col[rt] = c;
return ;
}
pushdown(rt);
mid;
if(L <= m) update(L,R,c,lson);
if(R > m) update(L,R,c,rson);
} void query(int l,int r,int rt){
if(col[rt]!=-){
if(Hash[col[rt]]==) cnt++;
Hash[col[rt]] = ;
return ;
}
if(l == r) return;
mid;
query(lson);
query(rson);
} int bin(int key,int n,int a[]){
int l = ,r = n-;
while(l <= r){
mid;
if(a[m] == key) return m;
else if(a[m] < key) l = m+;
else r = m-;
}
return -;
}
int main()
{
ios::sync_with_stdio();
cin.tie(); cout.tie();
int t,n;
cin>>t;
while(t--){
cin>>n;
memset(col,-,sizeof(col));
memset(Hash,,sizeof(Hash));
int nn = ;
cnt = ;
for(int i = ;i < n;i++){
cin>>l[i]>>r[i];
a[nn++] = l[i];a[nn++] = r[i];
}
sort(a,a+nn);
int m = ;
for(int i = ;i < nn;i ++){
if(a[i]!=a[i-]) a[m++] = a[i];
}
for(int i = m-;i > ;i --){
if(a[i]!=a[i-]+) a[m++] = a[i] + ;
}
sort(a,a+m);
for(int i = ;i < n;i ++){
int li = bin(l[i],m,a);
int ri = bin(r[i],m,a);
update(li,ri,i,,m,);
}
query(,m,);
cout<<cnt<<endl;
}
}
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