zoj 1610 Count the Colors 线段树，成段更新染色

Count the ColorsTime Limit:2000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

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Description

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：对一段区间进行涂色，问最后每一种可见的颜色共占多少个不连续的区间。

方法：线段树+lazy标志

思路：这到题目的关键一点是区间不能取闭区间，根据题意，为了方便可以取左开右闭，而且要搞明白涂色的是区间而不是点，因此根据线段树不能一步得到结果，我们可以利用线段树的延时更新策略，即一个节点表示一段区间，如果这段区间是纯色的，那么会在这个点上进行标记，于是就不需要对其子区间进行图色了，除非下次图色的时候发生覆盖，才需要继续向下更新，最终线段树可以以较小的代价得到最终的色彩分布，然后，我们再去线性扫描一遍这个分布，就可以得到每种颜色的分布，将颜色值作为数组的下标，这样在扫描完成后得到的数据就是按照颜色值的变化的。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define maxn 8010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;

int color[maxn<<2];
int distrb[maxn];//颜色分布
int res[maxn];

void push_down(int rt)
{
    if(color[rt]!=-1)//如果有延时更新标志
    {
        color[rt<<1]=color[rt<<1|1]=color[rt];//更新子节点的标志
        color[rt]=-1;//更新父节点的标志
    }
}

void update(int L,int R,int data,int l,int r,int rt)
{
    if(L<=l&&R>=r)//找到区间
    {
        color[rt]=data;
        return;
    }
    if(color[rt]==data)//如果该区间包含该标记
        return;
    push_down(rt);//向下更新
    int m=(l+r)>>1;
    if(L<=m)
        update(L,R,data,lson);
    if(R>m)
        update(L,R,data,rson);
}

void query(int l,int r,int rt)
{
    if(color[rt]!=-1)//该区间是纯色的
    {
        for(int i=l; i<=r; i++)
            distrb[i]=color[rt];//将颜色的分布还原
        return;
    }
    if(l!=r&&color[rt]==-1)//只有区间长度大于0才是有意义的
    {
        int m=(l+r)>>1;
        query(lson);
        query(rson);
    }

}



int main()
{
    int n,i;
    int left,right,data;
    int temp;
    while(~scanf("%d",&n))
    {
        memset(color,-1,sizeof(color));
        memset(distrb,-1,sizeof(distrb));
        while(n--)
        {
            scanf("%d%d%d",&left,&right,&data);
            update(left+1,right,data,1,maxn,1);//区间是左开右闭的
        }
        query(1,maxn,1);//查询整个区间
        memset(res,0,sizeof(res));
        for(i=0;i<maxn;)//遍历颜色分布，统计不同的色块
        {
            while(i<maxn&&distrb[i]==-1)
                i++;
            if(i>=maxn)
                break;
            temp=distrb[i];
            res[temp]++;
            while(i<maxn&&distrb[i]==temp)
                i++;
        }
        for(i=0;i<maxn;i++)//输出
        {
            if(res[i]!=0)
                printf("%d %d\n",i,res[i]);
        }
        printf("\n");
    }
    return 0;
}