POJ 2528 Mayor's posters（线段树）

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.

 

题目大意：给n个区间，每个区间上一种颜色（会覆盖掉之前的），问最终还有多少种颜色在上面。

思路：线段树水题，不多说了。先离散化，详细看代码，可能方法比较奇葩……最后还有一点要注意的是区间[l, r]是指第l条线段到第r条线段，而不是第l个点到第r个点。

 

代码（63MS）：

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 const int MAXN = 10010;

 8 

 9 struct Node {

10     int x, id;

11     bool operator < (const Node &rhs) const {

12         return x < rhs.x;

13     }

14 };

15 

16 int tree[MAXN * 4];

17 Node a[MAXN * 2];

18 int l[MAXN], r[MAXN];

19 int n, cnt, m;

20 

21 void pushdown(int x) {

22     if(!tree[x]) return ;

23     int ll = x << 1, rr = ll ^ 1;

24     tree[ll] = tree[rr] = tree[x];

25     tree[x] = 0;

26 }

27 

28 void update(int x, int left, int right, int L, int R, int key) {

29     if(L <= left && right <= R) {

30         tree[x] = key;

31     }

32     else {

33         pushdown(x);

34         int ll = x << 1, rr = ll ^ 1;

35         int mid = (left + right) >> 1;

36         if(L < mid) update(ll, left, mid, L, R, key);

37         if(mid < R) update(rr, mid, right, L, R, key);

38     }

39 }

40 

41 bool vis[MAXN];

42 

43 int query(int x, int left, int right) {

44     int ret = 0;

45     if(tree[x]) {

46         if(!vis[tree[x]]) {

47             vis[tree[x]] = true;

48             ret = 1;

49         }

50     }

51     else {

52         if(right - left == 1) return 0;

53         int ll = x << 1, rr = ll ^ 1;

54         int mid = (left + right) >> 1;

55         ret += query(ll, left, mid);

56         ret += query(rr, mid, right);

57     }

58     return ret;

59 }

60 

61 int main() {

62     int T;

63     scanf("%d", &T);

64     while(T--) {

65         scanf("%d", &n);

66         cnt = 0;

67         for(int i = 0; i < n; ++i) {

68             scanf("%d%d", &l[i], &r[i]);

69             a[cnt].x = l[i]; a[cnt++].id = i;

70             a[cnt].x = r[i] + 1; a[cnt++].id = i + MAXN;

71         }

72         sort(a, a + cnt);

73         m = 1;

74         for(int i = 0; i < cnt; ++i) {

75             if(i > 0 && a[i - 1].x != a[i].x) ++m;

76             if(a[i].id < MAXN) l[a[i].id] = m;

77             else r[a[i].id - MAXN] = m;

78         }

79         memset(tree, 0, sizeof(tree));

80         for(int i = 0; i < n; ++i)

81             update(1, 1, m, l[i], r[i], i + 1);

82         memset(vis, 0, sizeof(vis));

83         int ans = query(1, 1, m);

84         printf("%d\n", ans);

85     }

86 }

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