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Mayor's posters
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 56864 | Accepted: 16445 |
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.
- 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
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.
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
第一遍TLE代码，至今还未找出错误；
（还小啊~~）
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define N 1005 int m,ans,li[N],ri[N]; int x[N<<3],col[N<<4],hash[N]; void pushdown(int rt){ col[rt<<1]=col[rt<<1|1]=col[rt]; col[rt]=-1; } void updata(int rt,int l,int r,int val,int x,int y){ if(x>=l&&y<=r){ col[rt]=val;return ; } if(col[rt]!=-1) pushdown(rt); int mid=(x+y)>>1; if(mid>=l) updata(rt<<1,l,r,val,x,mid); if(mid<r) updata(rt<<1|1,l,r,val,mid+1,y); } void query(int rt,int l,int r){ if(l==r){ if(!hash[col[rt]]){ hash[col[rt]]=1; ans++; } return ; } if(col[rt]!=-1) pushdown(rt); int mid=(l+r)>>1; query(rt<<1,l,mid); query(rt<<1|1,mid+1,r); } int search(int ll,int hh,int xx){ while(ll<=hh){ int mid=(ll+hh)>>1; if(x[mid]==xx) return mid; else if(x[mid]>xx) hh=mid-1; else ll=mid+1; } return -1; } int main(){ int t,n; scanf("%d",&t); while(t--){ memset(col,-1,sizeof col); memset(hash,0,sizeof hash); int tot(0); scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d%d",&li[i],&ri[i]); x[++tot]=li[i]; x[++tot]=ri[i]; } sort(x+1,x+tot+1); m=1; for(int i=2;i<=tot;i++){ if(x[i]!=x[i-1]) x[++m]=x[i]; } for(int i=m;i>1;i--){ if(x[i]-x[i-1]>1) x[++m]=x[i]-1; } sort(x+1,x+m+1); for(int i=1;i<=n;i++){ int l=search(1,m,li[i]); int r=search(1,m,ri[i]); updata(1,l,r,i,1,m); } ans=0; query(1,1,m); printf("%d\n",ans); } return 0; }
题意:在墙上贴海报,海报可以互相覆盖,问最后可以看见几张海报
思路:这题数据范围很大,直接搞超时+超内存:
需要：线段树2的板子+离散化
离散化，如下面的例子（题目的样例），因为单位1是一个单位长度，将下面的
1 2 3 4 6 7 8 10
— — — — — — — —
1 2 3 4 5 6 7 8
离散化 X[1] = 1; X[2] = 2; X[3] = 3; X[4] = 4; X[5] = 6; X[7] = 8; X[8] = 10
于是将一个很大的区间映射到一个较小的区间之中了，然后再对每一张海报依次更新在宽度为1~8的墙上(用线段树），最后统计不同颜色的段数。
但是只是这样简单的离散化是错误的，
如三张海报为：1~10 1~4 6~10
离散化时 X[ 1 ] = 1, X[ 2 ] = 4, X[ 3 ] = 6, X[ 4 ] = 10
第一张海报时：墙的1~4被染为1；
第二张海报时：墙的1~2被染为2，3~4仍为1；
第三张海报时：墙的3~4被染为3，1~2仍为2。
最终，第一张海报就显示被完全覆盖了，于是输出2，但实际上明显不是这样，正确输出为3。
新的离散方法为：在相差大于1的数间加一个数，例如在上面1 4 6 10中间加5（算法中实际上1，4之间，6，10之间都新增了数的）
X[ 1 ] = 1, X[ 2 ] = 4, X[ 3 ] = 5, X[ 4 ] = 6， X[ 5 ] = 10
这样之后，第一次是1~5被染成1；第二次1~2被染成2；第三次4~5被染成3
最终，1~2为2，3为1，4~5为3，于是输出正确结果3。
所以离散化要保存所有需要用到的值,排序后,分别映射到1~n,这样复杂度就会小很多很多
而这题的难点在于每个数字其实表示的是一个单位长度(并且一个点),这样普通的离散化会造成许多错误(包括我以前的代码,poj这题数据奇弱)
给出下面两个简单的例子应该能体现普通离散化的缺陷:
1-10 1-4 5-10
1-10 1-4 6-10
为了解决这种缺陷,我们可以在排序后的数组上加些处理,比如说[1,2,6,10]
如果相邻数字间距大于1的话,在其中加上任意一个数字,比如加成[1,2,3,6,7,10],然后再做线段树就好了.
线段树功能:update:成段替换 query:简单hash
AC代码：
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define N 10005 int m,li[N],ri[N]; int x[N<<3],col[N<<4],ans; bool hash[N]; void pushdown(int rt) { col[rt<<1]=col[rt<<1|1]=col[rt]; col[rt]=-1; } void updata(int rt,int L,int R,int c,int l,int r){ if(l>=L&&r<=R){ col[rt]=c;return ; } if(col[rt]!=-1) pushdown(rt); int mid=(l+r)>>1; if(mid>=L) updata(rt<<1,L,R,c,l,mid); if(mid<R) updata(rt<<1|1,L,R,c,mid+1,r); } void query(int rt,int l,int r){ if(l==r){ if(!hash[col[rt]]){ hash[col[rt]]=1; ans++; } return; } if(col[rt]!=-1) pushdown(rt); int mid=(l+r)>>1; query(rt<<1,l,mid); query(rt<<1|1,mid+1,r); } int search(int ll,int hh,int xx){ int mid; while (ll<=hh){ mid=(ll+hh)>>1; if(x[mid]==xx) return mid; else if(x[mid]>xx) hh=mid-1; else ll=mid+1; } return -1; } int main(){ int t,n,i; scanf("%d",&t); while(t--){ memset(col,-1,sizeof(col)); memset(hash,0,sizeof(hash)); int tot=0; scanf("%d",&n); for(i=1;i<=n;i++){ scanf ("%d %d",&li[i],&ri[i]); x[++tot]=li[i]; x[++tot]=ri[i]; } sort(x+1,x+tot+1); m=1; for(i=2;i<=tot;i++){ if(x[i]!=x[i-1]) x[++m]=x[i]; } for(i=m;i>1;i--){ if(x[i]-x[i-1]>1) x[++m]=x[i]-1; } sort(x+1,x+m+1); for(i=1; i<=n; i++){ int l=search(1,m,li[i]); int r=search(1,m,ri[i]); updata(1,l,r,i,1,m); } ans=0; query(1,1,m); printf("%d\n",ans); } return 0; }
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