树状数组 POJ 2481 Cows

 

题目传送门

 1 #include <cstdio>  2 #include <cstring>  3 #include <algorithm>  4 using namespace std;  5  6 const int MAX_N = 100000 + 10;  7 int cnt[MAX_N];  8 int ans[MAX_N];  9 int maxn = -1;  10 struct node  11 {  12 int s, e;  13 int id;  14 }cow[MAX_N];  15  16 inline int read(void)  17 {  18 int x = 0, f = 1; char ch = getchar ();  19 while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar ();}  20 while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar ();}  21 return x * f;  22 }  23  24 inline void print(int x)  25 {  26 if (x < 0) {putchar ('-'); x = -x;}  27 if (x > 9) print (x / 10);  28 putchar (x % 10 + '0');  29 }  30  31 bool cmp(node x, node y) //先对e从大到小排序  32 {  33 if (x.e == y.e)  34 return x.s < y.s; //如果如果相等，再按s从小到大排序  35 else  36 return x.e > y.e;  37 }  38  39 int lowbit(int t) //求 2^t  40 {  41 //return t & (t ^ (t - 1));  42 return t & (-t);  43 }  44  45 void plus(int x)  46 {  47 while (x <= maxn)  48  {  49 ans[x]++; //记录强壮的牛数  50 x += lowbit (x);  51  }  52 }  53  54 int sum(int x) //统计前x项强壮的牛数  55 {  56 int sum = 0;  57 while (x > 0)  58  {  59 sum += ans[x];  60 x -= lowbit(x);  61  }  62 return sum;  63 }  64  65 int main(void) //POJ 2481 Cows  66 {  67 //freopen ("inH.txt", "r", stdin);  68 int n;  69  70 while (scanf ("%d", &n) && n)  71  {  72 for (int i=1; i<=n; ++i)  73  {  74 cow[i].s = read (); cow[i].e = read (); //读入外挂优化  75 //scanf ("%d%d", &cow[i].s, &cow[i].e);  76 cow[i].id = i;  77 maxn = max (maxn, cow[i].e);  78  }  79 memset (ans, 0, sizeof (ans));  80 sort (cow+1, cow+n+1, cmp);  81 for (int i=1; i<=n; ++i)  82  {  83 if (cow[i].e == cow[i-1].e && cow[i].s == cow[i-1].s) //如果值相等，不加入统计  84 cnt[cow[i].id] = cnt[cow[i-1].id];  85 else  86  {  87 cnt[cow[i].id] = sum (cow[i].s + 1); //第id的牛有多少比它强壮  88  }  89 plus (cow[i].s + 1);  90  }  91 int j;  92 for (j=1; j<n; ++j)  93  {  94 print (cnt[j]); putchar (' '); //输出外挂优化  95 //printf ("%d ", cnt[j]);   96  }  97 printf ("%d\n", cnt[j]);  98  }  99 100 return 0; 101 }