ZOJ 1610 Count the Colors（線段樹(shù)lazy+暴力統(tǒng)計(jì)）
來(lái)源：程序員人生 發(fā)布時(shí)間：2015-03-06 09:03:20 閱讀次數(shù)：2777次
Count the Colors
Time Limit: 2 Seconds Memory Limit: 65536 KB
Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.
Your task is counting the segments of different colors you can see at last.
Input
The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.
Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:
x1 x2 c
x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.
All the numbers are in the range [0, 8000], and they are all integers.
Input may contain several data set, process to the end of file.
Output
Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.
If some color can't be seen, you shouldn't print it.
Print a blank line after every dataset.
Sample Input
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
具體做法就是區(qū)間更新的時(shí)候lazy操作，端點(diǎn)統(tǒng)計(jì)的時(shí)候暴力，。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn=8000+10;
int sum[maxn<<2];
int cnt,n;
int col[maxn],ans[maxn];
void pushdown(int rs)
{
if(sum[rs]!=⑴)
{
sum[rs<<1]=sum[rs<<1|1]=sum[rs];
sum[rs]=⑴;
}
}
void update(int x,int y,int c,int l,int r,int rs)
{
if(l>=x&&r<=y)
{
sum[rs]=c;
return ;
}
pushdown(rs);
int mid=(l+r)>>1;
if(x<=mid) update(x,y,c,l,mid,rs<<1);
if(y>mid) update(x,y,c,mid+1,r,rs<<1|1);
}
void solve(int l,int r,int rs)
{
if(l==r)
{
col[cnt++]=sum[rs];
return ;
}
pushdown(rs);
int mid=(l+r)>>1;
solve(l,mid,rs<<1);
solve(mid+1,r,rs<<1|1);
}
int main()
{
int l,r,x;
while(~scanf("%d",&n))
{
cnt=0;
CLEAR(sum,⑴);
CLEAR(ans,0);
int m=8000;
REP(i,n)
{
scanf("%d%d%d",&l,&r,&x);
update(l,r⑴,x,0,m,1);
}
solve(0,m,1);
REP(i,cnt)
{
int j=i+1;
if(col[i]!=⑴) ans[col[i]]++;
while(col[j]==col[i]&&j<cnt)
j++;
i=j⑴;
}
for(int i=0;i<=maxn;i++)//for色彩種類
if(ans[i]) printf("%d %d
",i,ans[i]);
puts("");
}
return 0;
}
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