POJ 2528 线段树的离散化 和坐标位置的判定

Mayor's posters

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 55154 Accepted: 15999

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 

• 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

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l_i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l_i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l_i, l_i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 

Sample Input

151 42 68 103 47 10

Sample Output

4

题目大意：

贴n张图片，后一张会把前面一张给覆盖中，问，最后能看到的能有几张

思路：

由于坐标过大，那么如果开数组肯定胡超过这个区间。那么我们就要思考着离散化。离散化的步骤是首先用一个数组保存这些数据，然后去重，然后用lowerbound来寻找他的坐标并保存就可以了。

然后对于题目的这种情况，我们考虑到的是先update，然后用tree数组来保存这个坐标值，但是注意，每次如果找到了这个tree[o]要记得往下传值。

然后还有一个注意的一点就是为了防止一个点重复的ans++，所以就要用一个atlas数组来保存已经访问过的点的信息就可以了。

然后query的任务就是访问所有的段。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int maxn = 10000 + 5;int n;int pos[maxn][2];int tree[maxn << 2];int a[maxn * 4];int atlas[maxn * 4];ll ans;int al;void init(){ memset(pos, 0, sizeof(pos)); memset(a, 0, sizeof(a)); memset(tree, 0, sizeof(tree)); memset(atlas, 0, sizeof(atlas)); scanf("%d", &n); al = 0; for (int i = 1; i <= n; i++){ scanf("%d%d", &pos[i][0], &pos[i][1]); a[++al] = pos[i][0], a[++al] = pos[i][1]; } /* for (int i = 1; i <= al; i++){ printf("%d%c", a[i], i == al ? '\n' : ' '); } */}void pushdown(int o){ if (tree[o] == 0) return ; int l = o << 1; int r = o << 1 | 1; tree[l] = tree[r] = tree[o]; tree[o] = 0;}void update(int o, int ul, int ur, int l, int r, int pos){ if (ul <= l && ur >= r){ tree[o] = pos; return ; } pushdown(o); int mid = l + (r - l) / 2; if (ul <= mid){ update(o << 1, ul, ur, l, mid, pos); } if (ur > mid){ update(o << 1 | 1, ul, ur, mid + 1, r, pos); }}void query(int o, int l, int r){ if (tree[o] != 0){ if (atlas[tree[o]] == 0) { ans++; atlas[tree[o]] = 1; } tree[o] = 0; return ; } int mid = l + (r - l) / 2; query(o << 1, l, mid); query(o << 1 | 1, mid + 1, r);}void solve(){ sort(a + 1, a + 1 + al); int cnt = 1; for (int i = 2; i <= al; i++){ if (a[i] == a[i - 1]) continue; a[++cnt] = a[i]; } /* for (int i = 1; i <= cnt; i++){ printf("%d%c", a[i], i == cnt ? '\n' : ' ' ); } */ ans = 0; for (int i = 1; i <= n; i++){ int ul = lower_bound(a + 1, a + 1 + cnt, pos[i][0]) - a; int ur = lower_bound(a + 1, a + 1 + cnt, pos[i][1]) - a; pos[i][0] = ul, pos[i][1] = ur; update(1, ul, ur, 1, cnt, i); } query(1, 1, cnt); printf("%I64d\n", ans);}int main(){ int t; scanf("%d", &t); while (t--){ init(); solve(); } return 0;}</cstring></algorithm></cstdio>