POJ 2528 Mayor’s posters
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 37982 | Accepted: 11030 |
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
题目意思是在墙上贴海报,海报可以相互覆盖,问你最后可以看见几张海报。
因为数据量大，若不离散化会超时或超内存。故学习了别人写法，简单的hash后再线段树成段更新。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define maxn 10010
bool hash[maxn];
int li[maxn] , ri[maxn];
int X[maxn<<4];
int col[maxn<<4];
int cnt;
void PushDown(int rt)
{
if (col[rt] != -1)
{
col[rt<<1] = col[rt<<1|1] = col[rt];
col[rt] = -1;
}
}
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);
int m =(l+r)>> 1;
if (L<=m) update(L , R , c , lson);
if (m<R) update(L , R , c , rson);
}
void query(int l,int r,int rt)
{
if (col[rt] != -1)
{
if (!hash[col[rt]]) cnt ++;
hash[ col[rt] ] = true;
return ;
}
if (l == r) return ;
int m = (l + r)>>1;
query(lson);
query(rson);
}
int Bin(int key,int n,int X[])
{
int l = 0 , r = n - 1;
while (l <= r)
{
int m = (l + r) >> 1;
if (X[m] == key) return m;
if (X[m] < key) l = m + 1;
else r = m - 1;
}
return -1;
}
int main()
{
int T , n;
scanf("%d",&T);
while (T --)
{
scanf("%d",&n);
int nn = 0;
for (int i=0; i<n; i ++)
{
scanf("%d%d",&li[i],&ri[i]);
X[nn++] = li[i];
X[nn++] = ri[i];
}
sort(X , X+nn);
int m = 1;
for (int i = 1 ; i < nn; i ++)
{
if (X[i] != X[i-1]) X[m ++] = X[i];
}
for (int i = m - 1 ; i > 0 ; i --)
{
if (X[i] != X[i-1] + 1) X[m ++] = X[i-1] + 1;
}
sort(X , X + m);
memset(col , -1 , sizeof(col));
for (int i = 0 ; i < n ; i ++)
{
int l = Bin(li[i], m, X);
int r = Bin(ri[i], m, X);
update(l, r, i, 0, m - 1, 1);
}
cnt = 0;
memset(hash, false, sizeof(hash));
query(0, m - 1, 1);
printf("%d\n",cnt);
}
return 0;
}