传统_POJ-2528

题目链接：http://poj.org/problem?id=2528

搞一个简单的映射就可以了，代码如下

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cstring>

 4 #include<algorithm>

 5 using namespace std;

 6 #define lowbit(a) ((a)&(-a))

 7 #define max(a, b) ((a)>(b)?(a):(b))

 8 #define min(a, b) ((a)<(b)?(a):(b))

 9 #define MAXN 10010

10 #define PI 3.1415926

11 #define E 2.718281828

12 #define INF 0x777777f

13 typedef long long LL;

14 

15 struct heap{

16     #define hMAXN 10010

17     int hCnt, a[hMAXN], idx[hMAXN][2];

18     inline void clear(){hCnt = 0;}

19     inline void swp(int p, int q){a[q] = a[p]; idx[q][0] = idx[p][0];}

20     inline void maintain(int p){idx[idx[p][0]][1] = p;}

21     void swim(int p){

22         int tmp[] = {a[p], idx[p][0]};

23         for (int q = p >> 1; q > 0; p = q, q >>= 1){

24             if (a[q] > tmp[0]){

25                 swp(q, p); maintain(p);

26             } else break;

27         } a[p] = tmp[0]; idx[p][0] = tmp[1]; maintain(p);

28     }

29     void sink(int p){

30         int tmp[] = {a[p], idx[p][0]};

31         for (int q = p << 1; q <= hCnt; p = q, q <<= 1){

32             if (q < hCnt && a[q] > a[q+1]) q++;

33             if (a[q] < tmp[0]){

34                 swp(q, p); maintain(p);

35             } else break;

36         } a[p] = tmp[0]; idx[p][0] = tmp[1]; maintain(p);

37     }

38     void push(int x){a[++hCnt] = x; swim(hCnt);}

39     void push(int x, int p){a[++hCnt] = x; idx[hCnt][0] = p; swim(hCnt);}

40     int pop(int p){int ret = a[p]; swp(hCnt, p); a[hCnt] = idx[hCnt][0] = 0; hCnt--; swim(p); sink(p); return ret;}

41 } h;

42 struct node{

43     int a, b, c;

44 } x[MAXN*2];

45 int n, vis[MAXN];

46 

47 bool cmp(node a, node b){return (a.a<b.a)||(a.a==b.a)&&(a.b<b.b);}

48 void init(){

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

50     for (int i = 1; i <= n; i++){

51         vis[i] = 0;

52         scanf("%d%d", &x[i*2-1].a, &x[i*2].a); x[i*2].a++;

53         x[i*2-1].b = 0; x[i*2-1].c = i;

54         x[i*2].b = 1;    x[i*2].c = i;

55     }

56     sort(x+1, x+1+2*n, cmp);

57     h.clear();

58 }

59 void work(){

60     for (int i = 1; i < 2*n; i++){

61         for ( ; x[i].a==x[i+1].a; i++){

62             if (!x[i].b){h.push(-x[i].c, x[i].c);}

63             else {h.pop(h.idx[x[i].c][1]);}

64         }

65         if (!x[i].b){h.push(-x[i].c, x[i].c);}

66         else {h.pop(h.idx[x[i].c][1]);}

67         if (h.hCnt) vis[-h.a[1]] = 1;

68     }

69     int ans = 0;

70     for (int i = 1; i <= n; i++)

71         if (vis[i]) ans++;

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

73 }

74 int main(){

75     int t; scanf("%d", &t);

76     for (int i = 0; i < t; i++){

77         init();

78         work();

79     }

80     return 0;

81 }