POJ Mayor’s posters
描述
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.
输入
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.
输出
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.
样例输入
1 5 1 4 2 6 8 10 3 4 7 10
样例输出
4
来源Alberta Collegiate Programming Contest 2003.10.18
md这个好坑啊，调了好久才发现离散化有锅
然后发现网上找的代码基本全锅了，导致对拍都没看出问题（雾
最后思索了一下，发现我锅的方式正好和他们的相反qwq
题目应该还是比较简单的，离散化一下，然后按顺序打标记用线段树维护一下，最后查一遍最后哪些标号出现就好了
这个离散化当时没怎么思考就无脑随便写了下（没有严格贴近的那种qwq，最后才发似乎有点锅w
我们发现这个在边界维护是需要注意的，就如（1，4）（1，2）（3，4）这时候后两个是将第一个覆盖掉的，因此我们离散化之后2、3也是应该贴近的，这样才能保证合法性。
而另一组（1，5）（1，2）（4，5），此时后两个并没有将其覆盖，即我们发现在离散化之前存在间隔离散化之后一定需要保持间隔，否则就会被误判称上一组情况（网上大部分代码应该都没判这里qwq）
因此，我们在离散化时候同样要保持其间隔性，即对于两个相近节点如果离散化之前存在间隔，在离散化时候也要插入空白间隔维持信息。
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define inf 1000000007
#define ll long long
#define N 40010
inline int rd()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int T,n,l[N],r[N],tq[N];
int lz[N<<2],c[N<<2];
void pd(int l,int r,int p){
if(!lz[p]) return;
lz[p<<1]=lz[p<<1|1]=lz[p];
lz[p]=0;
}
void cg(int l,int r,int p,int L,int R,int v){
if(l==L&&r==R){
lz[p]=v;
return;
}
pd(l,r,p);
int mid=(l+r)>>1;
if(R<=mid) cg(l,mid,p<<1,L,R,v);
else if(L>mid) cg(mid+1,r,p<<1|1,L,R,v);
else{
cg(l,mid,p<<1,L,mid,v);
cg(mid+1,r,p<<1|1,mid+1,R,v);
}
}
int fd(int l,int r,int p,int x){
if(l==r) return lz[p];
pd(l,r,p);
int mid=(l+r)>>1;
return (x<=mid?fd(l,mid,p<<1,x):fd(mid+1,r,p<<1|1,x));
}
bool vs[N];
void work(){
memset(vs,0,sizeof(vs));
memset(lz,0,sizeof(lz));
n=rd();
for(int i=1;i<=n;++i){
l[i]=rd();r[i]=rd();
tq[i*2-1]=l[i];
tq[i<<1]=r[i];
}
sort(tq+1,tq+n+n+1);
map<int,int>mp;
for(int i=1,j=0;i<=(n<<2);++i){
if(i==1||tq[i]!=tq[i-1]){
++j;
if(i^1&&tq[i]-tq[i-1]>1) ++j;
mp[tq[i]]=j;
}
}
for(int i=1;i<=n;++i){
cg(1,n<<2,1,mp[l[i]],mp[r[i]],i);
}
int ans=0;
for(int i=1,x;i<=(n<<2);++i){
x=fd(1,n<<2,1,i);
if(x&&!vs[x]) vs[x]=1,++ans;
}
printf("%d\n",ans);
}
int main()
{
T=rd();
while(T--) work();
return 0;
}
发表评论