Count the Colors

最新推荐文章于 2018-08-11 10:39:01 发布

原创最新推荐文章于 2018-08-11 10:39:01 发布 · 206 阅读

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本文详细解析了一种区间染色算法，旨在解决给定区间内多次染色操作后，每种颜色可见线段数量的问题。通过构建线段树并进行区间更新与查询，实现了高效的颜色计数。

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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

题意：给一个区间，有n次染色操作，每次将[x1, x2]染为c，求最后每种颜色各有多少线段可以看到。

#include <iostream>
#include <bits/stdc++.h>
#define maxn 8010
using namespace std;
int color[maxn<<2];
int ans[maxn];
int last,n;//last用于保存上一个节点涂的颜色
void pushdown(int rt)
{
    if(color[rt]!=-1){
        color[rt<<1]=color[rt<<1|1]=color[rt];
        color[rt]=-1;
    }
    return ;
}
void update(int rt,int L,int R,int l,int r,int c)
{
    if(L<=l&&R>=r){
        color[rt]=c;
        return ;
    }
    if(color[rt]==c) return ;
    int m=(l+r)>>1;
    pushdown(rt);
    if(L<=m) update(rt<<1,L,R,l,m,c);
    if(m<R) update(rt<<1|1,L,R,m+1,r,c);
}
void query(int l,int r,int rt)
{
    if(l==r){
        if(color[rt]!=-1&&color[rt]!=last){
            ans[color[rt]]++;
        }
        last=color[rt];
        return ;
    }
    if(l==r) return ;
    pushdown(rt);//标记下推
    int m=(l+r)>>1;
    query(l,m,rt<<1);
    query(m+1,r,rt<<1|1);
}
int main()
{
    int n;
    int l,r,c;
    while(scanf("%d",&n)!=EOF){
            fill(color,color+(maxn<<2),-1);
        for(int e=1;e<=n;e++){
            scanf("%d %d %d",&l,&r,&c);
            update(1,l+1,r,1,8000,c);
        }
        last=-1;
        memset(ans,0,sizeof(ans));
        query(1,8000,1);
        for(int i=0;i<=8000;i++){
            if(ans[i]!=0){
                printf("%d %d\n",i,ans[i]);
            }
        }
        printf("\n");
    }
    return 0;
}