Poj 2528 Mayor & amp; #39; s posters (dynamic line segment tree)

Poj 2528 Mayor & #39; s posters (dynamic line segment tree)

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Question:

Given ~ The range of 10000000, and then there are N operations (N <= 10000), the I operation is ~ The r range covers I. How many colors are there in the last one.

Solution:

At first, I thought about discretization, but I thought it was a little troublesome. Then I asked my teammates who worked on data structures. Then he talked about the dynamic line segment tree. The idea is as follows:

Define an ID. Then, the root node 1 indicates the period in charge of the 1-MAXN color. Then every time, it is a dynamic building. When the left subinterval of an interval does not exist. Create it and record the ID of the Left subinterval and right subinterval of each interval.

Finally, we use a dfs set to record the total number of colors.

Note:

When the left subinterval and right subinterval of an interval are updated by it, it must be cleared.

And the maxn. The START is unknown.

#include 
 
  #include 
  
   using namespace std;#define maxn 2000000int lson[maxn],rson[maxn],color[maxn];int cnt,root;void build(){    cnt = 2;    root = 1;    lson[root] = 0;    rson[root] = 0;    color[root] = 0;}void pushdown(int id){    if(!lson[id])    {        lson[id] = cnt++;        lson[lson[id]] = 0;        rson[lson[id]] = 0;        color[lson[id]] = 0;    }    if(!rson[id])    {        rson[id] = cnt++;        lson[rson[id]] = 0;        rson[rson[id]] = 0;        color[rson[id]] = 0;    }    if(color[id])    {        color[lson[id]] = color[id];        color[rson[id]] = color[id];        color[id] = 0;    }}void op(int id,int ls,int rs,int l,int r,int c){    if(ls >= l && rs <= r)    {        color[id] = c;        return;    }    pushdown(id);    int mid = (ls+rs)>>1;    if(l <= mid) op(lson[id],ls,mid,l,r,c);    if(mid < r) op(rson[id],mid+1,rs,l,r,c);}set
   
     ans;void dfs(int id){    if(color[id])    {        ans.insert(color[id]);        return;    }    if(lson[id]) dfs(lson[id]);    if(rson[id]) dfs(rson[id]);}int main(){    int T;    scanf("%d",&T);    for(int ks = 1;ks <= T;ks++)    {        int n;        scanf("%d",&n);        ans.clear();        build();        for(int i = 1;i <= n;i++)        {            int l,r;            scanf("%d %d",&l,&r);            op(root,1,10000000,l,r,i);        }        dfs(root);        printf("%d\n",ans.size());    }    return 0;}