Click to open the link poj 2528
Description
- Every candidate can place exactly one poster on the wall.
- All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
- The wall is divided into segments and the width of each segment is one byte.
- Each poster must completely cover a contiguous number of wall segments.
They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections.
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall.
Input
Output
The picture below illustrates the case of the sample input.
Sample Input
1 5 1 4 2 6 8 10 3 4 7 10
Sample Output
4
Topic meaning:
It is to give you a poster. Posters will be posted in each section, and the ones posted at the back will cover the previous ones, so that you can ask how many posters you can see.
At the beginning, I thought that the interval merging of line segment trees was a problem, but this problem stuck me for a long time, because it needed to be discretized. I have been exposed to discretization before, but it is not commonly used, so I read other people's code, and I just looked at it at first. I don't understand, but I can still understand by outputting each step in the end.
The original ones are (1, 4), (2, 6), (8, 10), (3, 4), (7, 10). After discretization, these intervals are changed to (1, 4), ( 2, 5), (7, 8), (3, 4), (6, 8).
Code after a long time to complete:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define lson l,mid,ri<<1 #define rson mid +1 ,r,ri<<1|1 int n,cnt; const int maxn = 1e5+10; struct Node{ int l,r,n; } a[maxn<<2]; struct node{ int point,num; } s[maxn<<2]; int map[maxn<<1][2],ans,vis[maxn<<1]; int cmp(node x,node y){ return x.point<y.point; } void Build(int l,int r,int ri){ a[ri].l = l; a[ri].r = r; a[ri].n = 0; if (l == r){ return ; } int mid = (l+r)>>1; Build(lson); Build(rson); } void insert(int ri,int l,int r,int m){ if(a[ri].l == l && a[ri].r == r){ a[ri].n = m; return; } int mid = (a[ri].l+a[ri].r)>>1; if(a[ri].n>0){ a[ri<<1].n = a[ri<<1|1].n = a[ri].n; a[ri].n = 0; } if(l>=a[ri<<1|1].l) insert(ri<<1|1,l,r,m); else if(r<=a[ri<<1].r) insert(ri<<1,l,r,m); else{ insert(ri<<1,l,mid,m); insert(ri<<1|1,mid+1,r,m); } } void solve(int ri){ if(a[ri].n){ if(!vis[a[ri].n]){ ans++; vis[a[ri].n] = 1; } return; } solve(ri<<1); solve(ri<<1|1); return; } int main(){ int t; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i = 0; i < n; i++){ scanf("%d%d",&map[i][0],&map[i][1]); s[i<<1].point = map[i][0]; s[i<<1|1].point = map[i][1]; s[i<<1].num = -(i+1); s[i<<1|1].num = i+1; } // for (int i = 0; i < 2*n; i++){ // printf("%d %d %d\n",i,s[i].point,s[i].num); // } sort(s,s+2*n,cmp); // printf("\n"); // for (int i = 0; i < 2*n; i++){ // printf("%d %d %d\n",i,s[i].point,s[i].num); // } int temp = s[0].point,cnt = 1; for(int i = 0; i < 2*n; i++){ if(temp != s[i].point){ cnt++; temp = s[i].point; } if(s[i].num < 0) map[-s[i].num-1][0] = cnt; else map[s[i].num-1][1] = cnt; } Build(1,cnt,1); for(int i = 0; i < n; i++){ // printf("%d %d\n",map[i][0],map[i][1]); insert(1,map[i][0],map[i][1],i+1); } memset(vis,0,sizeof(vis)); years = 0; solve(1); printf("%d\n",ans); } return 0; }Work hard, come on!