Mayor‘s posters
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 49385 | Accepted: 14304 |
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:
- 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.
Sample Input
1 5 1 4 2 6 8 10 3 4 7 10
Sample Output
4
Source
Alberta Collegiate Programming Contest 2003.10.18
题目大意：在一面墙上贴海报，每张海报一定贴满一整张瓷砖（如上图）。问有几张海报能够被看到（露出一部分也算）。
分析：既然是区间上的问题，一定用线段树来求解。所以容易想到，以瓷砖作为区间，海报贴住瓷砖的范围就对区间进行覆盖。
问题抽象出来就是：在一面墙上染色，最后问有几种不同的颜色。
问题到这里已经差不多解决了。但是一看数据范围，瓷砖有10000000张，如果建树的话，肯定MLE了。这里就用到了离散化。
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离散化：通俗点说，离散化就是压缩区间，使原有的长区间映射到新的短区间，但是区间压缩前后的覆盖关系不变。举个例子：
有一条1到10的数轴（长度为9），给定4个区间[2,4] [3,6] [8,10] [6,9]，覆盖关系就是后者覆盖前者，每个区间染色依次为 1 2 3 4。
现在我们抽取这4个区间的8个端点，2 4 3 6 8 10 6 9
然后删除相同的端点，这里相同的端点为6，则剩下2 4 3 6 8 10 9
对其升序排序，得2 3 4 6 8 9 10
然后建立映射
2 3 4 6 8 9 10
↓ ↓ ↓ ↓ ↓ ↓ ↓
1 2 3 4 5 6 7
那么新的4个区间为 [1,3] [2,4] [5,7] [4,6]，覆盖关系没有被改变。新数轴为1到7，即原数轴的长度从9压缩到6，显然构造[1,7]的线段树比构造[1,10]的线段树更省空间，搜索也更快，但是求解的结果却是一致的。（以上我是复制别人的= = ，自己也是第一次接触离散化，只是觉得他讲的很通俗。）
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再回到这题，其实跟普通的离散化又是不一样的。原因看题目中的图：他一张瓷砖代表了一个区间，而不是一个点，这是离散化时需要处理的。
为什么如果代表一个区间的时候，就不能进行普通的离散化呢？
例如：对于区间【1,4】和【5,10】，如上图，【1,10】倍全部覆盖，如果对上述区间离散化，会得到【1,2】和【3,4】两个区间，可以认为【1,4】全部被覆盖。
在看【1,4】，【6,10】，离散化结果和【1,4】【5,10】一样，但实际上，【1,10】并没有被完全覆盖。
如何解决？对区间进行排序后遍历，如果相邻的两个数相差大于1，就随意加一个大小处在两个之中的数字。这样就很好的解决了上面的问题。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define M 10050
using namespace std;
struct post{
int l,r;
}post[M];
int x[M<<1],ans,a[M<<4];
int tohash[10000005];
struct tree{
int l,r,color;
}tree[1000005];
void build(int l,int r,int root)
{
tree[root].l=l;
tree[root].r=r;
tree[root].color=-1; //初始化赋为-1
if(l==r)return ;
int mid=l+r>>1;
build(l,mid,root<<1);
build(mid+1,r,root<<1|1);
}
void pushdown(int root){ //如果一个区间里面的颜色有多种，就认为color是-1
if(tree[root].color!=-1)
{tree[root<<1].color=tree[root<<1|1].color=tree[root].color;
tree[root].color=-1;
}
return;
}
void update(int l,int r,int z,int root)
{
if(l==tree[root].l&&tree[root].r==r){
tree[root].color=z;
return ;
}
pushdown(root);
int mid=tree[root].l+tree[root].r>>1;
if(r<=mid)update(l,r,z,root<<1);
else if(l>mid)update(l,r,z,root<<1|1);
else {
update(l,mid,z,root<<1);
update(mid+1,r,z,root<<1|1);
}
}
void query(int root)
{
int flag,j;
if(tree[root].color!=-1){ //如果当前区间不是多种颜色，就直接添加入数组。
a[ans++]=tree[root].color;
return;
}
if(tree[root].l==tree[root].r)return;
query(root<<1);
query(root<<1|1);
return;
}
int main()
{
int T,i,j,k,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
j=0;
for(i=0;i<n;i++)
{
scanf("%d%d",&post[i].l,&post[i].r);
x[j++]=post[i].l;
x[j++]=post[i].r;
}
sort(x,x+j);
j=unique(x,x+j)-x; //unique函数用于选出重复的数并放到数组最后，并返回不重复数列的末尾位置。（相当于将重复的数去掉了）
for(i=j-1;i>=0;i--)
if(x[i]!=x[i-1]+1)x[j++]=x[i-1]+1; //解决普通离散化的缺陷
sort(x,x+j);
int num=1;
for(i=0;i<j;i++)
tohash[x[i]]=num++; //离散化映射
build(1,num-1,1);
for(i=0;i<n;i++)
{
update(tohash[post[i].l],tohash[post[i].r],i+1,1);
}
memset(a,0,sizeof(a));
ans=0;
query(1);
sort(a,a+ans);
int y=1;
for(i=1;i<ans;i++)
if(a[i]!=a[i-1])y++; //这里将先前可能有重复的颜色去掉，只选不重复的颜色。
printf("%d\n",y);
}
return 0;
}
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poj 2528 Mayor's posters（线段树区间覆盖、离散化）