Zoj 1610 count the colors (line segment tree, segment update coloring)

Link:

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 1610.

Question:

Dye a line segment with a length of 8000, and dye the interval [a, B] into the C color each time. Obviously, the color after the link will overwrite the previous color.

Calculate the number of intervals on each color online segment after dyeing.

Analysis and Summary:

It is not difficult to update the lazy mark in segments, but it has been difficult for me for some reasons...

1. I did this question yesterday, but I have been connecting to the samples for a long time. = ~! When I got up in the morning, I suddenly realized that the problem was located. The dyeing section for each question was to dye [a, B] into C. I thought it was just to dye ~ All vertices of B are dyed as C. In fact, this is not the case. Instead, we need to dye not vertices, but intervals. For example, we need to dye [0, 1] instead of dyeing 0, 1, 2, instead, the unit interval [0, 1] is colored. Suppose there is an example:

1 2 1

3 4 1

So this example should output 1 2. Just draw a picture on the paper and we can see that the interval [2, 3] is not stained, so there are two intermittent intervals where the color is 1.

To solve this problem, build a line segment tree (8000,), representing the range of 1 ~ 8000, then update (8000, A + 1, B, c );

2. Because I didn't take a closer look at the question, I got a wrong place, so I 've been Re (segmentation fault on zoj ):

The first line of each data set contains exactly one integer N, 1 <= n <= 8000, equal to the number of colored segments.

This sentence means that only N line segments are colored, and I have always understood that it is colored on a line segment with N unit intervals ~ This is the tragedy. The positive solution is: all the numbers are in the range [0, 8000], and they are all integers.

Code:

#include<iostream>#include<cstdio>#include<cstring>#define mem(str,x) memset(str,(x),sizeof(str))#define FOR(i,s,t) for(int i=(s); i<(t); ++i)#define FF(i,n) for(int i=0; i<(n); ++i)#define mid ((left+right)>>1)#define len (right-left+1)#define lson rt<<1, left, m#define rson rt<<1|1, m+1, right#define STOP puts("Stop Here~");using namespace std;const int MAXN = 8005;int n,col[MAXN<<2],vis[MAXN<<2],ans[MAXN<<2];inline void push_down(int rt){    if(col[rt] != -1){        col[rt<<1] = col[rt<<1|1] = col[rt];        col[rt] = -1;    }}void update(int rt,int left,int right,int l,int r,int data){    if(l<=left && right<=r){        col[rt] = data;        return;    }    if(col[rt] == data) return;    if(col[rt]!=-1)push_down(rt);    int m = mid;    if(l <= m)update(lson,l,r,data);    if(r > m)update(rson,l,r,data);}void query(int rt,int left,int right){    if(col[rt]>=0){        for(int i=left; i<=right; ++i)            vis[i] = col[rt];        return;    }    if(left!=right && col[rt] == -1){        int m = mid;        query(lson);        query(rson);    }}int main(){    int a,b,c;    while(~scanf("%d",&n)){        memset(col,-1,sizeof(col));        for(int i=0; i<n; ++i){            scanf("%d%d%d",&a,&b,&c);            if(a>=b)continue;            update(1,1,8000,a+1,b,c);        }        mem(vis,-1);         query(1,1,8000);        int i = 1;        mem(ans,0);        while(i<MAXN){            int color=vis[i], j=i+1;            if(color==-1){++i; continue;}            while(vis[j]!=-1 && vis[j]==color && j<MAXN) ++j;            ++ans[color];            i=j;        }        for(int i=0; i<MAXN; ++i)if(ans[i])            printf("%d %d\n",i,ans[i]);        puts("");     }    return 0;}

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Original Http://blog.csdn.net/shuangde800,
D_double (reprinted please mark)