ZOJ 1610 Count the Colors（线段树lazy+暴力统计）

发布时间：2020-12-13 20:14:53 所属栏目：PHP教程来源：网络整理

导读：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

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

具体做法就是区间更新的时候lazy操作，端点统计的时候暴力，。

#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,mid+1,r,rs<<1|1); } void solve(int l,int rs) { if(l==r) { col[cnt++]=sum[rs]; return ; } pushdown(rs); int mid=(l+r)>>1; solve(l,rs<<1); solve(mid+1,rs<<1|1); } int main() { int l,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,m,1); } solve(0,1); 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色彩种类 if(ans[i]) printf("%d %d ",i,ans[i]); puts(""); } return 0; }

（编辑：李大同）

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