1003 line segment TREE tree array range coverage

Algorithm:

During the competition, the length of a line segment is sorted from large to small, and a query is inserted at a time. Always wa.

This algorithm proved wrong after the competition.

For example, the data set 1, 6, 5, 6, and 10 cannot be passed.

The correct algorithm should be to sort Line X from small to large, and Y from large to small.

Use a line segment tree or a tree array to maintain the range and availability.

View code

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<map>#include<iostream>#include<algorithm>#include<vector>using namespace std;struct node{  int left, right;  int num;  int lazy;}seg[4000010];int N, M;struct flower{  int a,b, c;  bool operator < (const flower &A) const  {    // if ( abs(A.b - A.a) != abs(b - a) )      //  return abs(A.b - A.a) < abs(b-a);     if( a != A.a )        return a < A.a;     else        return b > A.b;  }}p[201000];int v[400010];int an[400010];void build(int l, int r, int root){   int mid = (l + r) / 2;   seg[root].left = l;   seg[root].right = r;   seg[root].num = 0;   seg[root].lazy = 0;   if( l == r )   {      return;   }   build(l,mid,root*2);   build(mid+1,r,root*2+1);}void update(int x){   seg[x * 2].lazy += seg[x].lazy;   seg[x * 2].num +=  seg[x].lazy;   seg[x * 2 + 1].lazy += seg[x].lazy;   seg[x * 2 + 1].num += seg[x].lazy;   seg[x].lazy = 0;}void add( int root, int u, int v, int val){  int mid = (seg[root].left + seg[root].right) / 2;  if(  seg[root].left == u && v == seg[root].right )  {      seg[root].lazy += val;      seg[root].num +=  val;       return;  }    if( seg[root].lazy != 0 )    update( root );   if( u > mid )       add( root * 2 + 1, u, v, val);  else if( v <= mid )       add( root * 2, u, v, val);  else  {       add( root * 2, u, mid, val);      add( root * 2 + 1, mid + 1, v,val);      }    }int sum( int root,int u, int v){  int mid = (seg[root].left + seg[root].right) / 2;  if(  seg[root].left == u && seg[root].right == v )  {     return seg[root].num;      }    if( seg[root].lazy != 0 )      update(root);  if( u > mid )      return sum( root * 2 + 1, u, v);  else if( v <= mid )      return sum( root * 2, u, v);  else  {      return min(sum( root * 2, u, mid), sum( root * 2 + 1, mid + 1, v));   }    }int main( ){   int cnt;  while( scanf("%d",&N) != EOF )  {       cnt = 0;     for( int i = 1; i <= N; i++)     {        scanf("%d%d",&p[i].a,&p[i].b);        p[i].c = i;        v[cnt++] = p[i].a;        v[cnt++] = p[i].b;           }     sort(v, v + cnt);     sort(p + 1, p + N + 1);     cnt = unique(v, v + cnt) - v ;     build(0,cnt,1);     for( int i = 1; i <= N; i++)          an[i] = 0;     for( int i = 1; i <= N; i++)     {          if( i != 1 && p[i].a == p[i-1].a && p[i].b == p[i-1].b )        {            an[p[i].c] = an[p[i-1].c];            int x = lower_bound(v, v + cnt, p[i].a) - v;             int y = lower_bound(v, v + cnt, p[i].b) - v;             add(1, x, y, 1);            continue;        }        int x = lower_bound(v, v + cnt, p[i].a) - v;         int y = lower_bound(v, v + cnt, p[i].b) - v;        an[p[i].c] = sum(1, x, y);        add(1, x, y, 1);        }     printf("%d",an[1]);     for( int i = 2; i <= N; i++)     {           printf(" %d",an[i]);     }     puts("");  }  return 0;  }