线段树(单点更新) POJ 2828 Buy tickets

 

题目传送门

 1 /*  2  结点存储下面有几个空位  3  每次从根结点往下找找到该插入的位置，  4  同时更新每个节点的值  5 */  6 #include <cstdio>  7 #define lson l, m, rt << 1  8 #define rson m+1, r, rt << 1 | 1  9 10 const int MAX_N = 200000 + 10; 11 int pos[MAX_N], val[MAX_N]; 12 int sum[MAX_N << 2]; 13 int que[MAX_N]; 14 int id; 15 16 void build(int l, int r, int rt) 17 { 18 sum[rt] = r - l + 1; //节点有多少个空位置 19 if (l == r) return ; 20 int m = (l + r) >> 1; 21  build (lson); 22  build (rson); 23 } 24 25 void update(int p, int l, int r, int rt) 26 { 27 sum[rt]--; //插入一个减少一个空位置 28 if (l == r) //找到了传递该位置 29  { 30 id = l; 31 return ; 32  } 33 int m = (l + r) >> 1; 34 if (p <= sum[rt<<1]) 35  { 36  update (p, lson); 37  } 38 else 39  { 40 p -= sum[rt<<1]; //减去 41  update (p, rson); 42  } 43 } 44 45 int main(void) //POJ 2828 Buy tickets 46 { 47 //freopen ("inE.txt", "r", stdin); 48 int n; 49 50 while (~scanf ("%d", &n)) 51  { 52 build (1, n, 1); 53 for (int i=1; i<=n; ++i) 54  { 55 scanf ("%d%d", &pos[i], &val[i]); 56  } 57 for (int i=n; i>=1; --i) //最后插入的位置不变 58  { 59 update (pos[i]+1, 1, n, 1); 60 que[id] = val[i]; 61  } 62 printf ("%d", que[1]); 63 for (int i=2; i<=n; ++i) 64  { 65 printf (" %d", que[i]); 66  } 67 puts (""); 68  } 69 70 return 0; 71 }