POJ 2828Buy Tickets
题目大意是说有n个插入操作，每次把B插入到位置A，原来A以后的全部往后移动1,球最后的序列
tree里保存的应该是这整个区间还有多扫个位置可以插入数据，那么线段树里从后往前扫描依次插入数据
比如现在吧B插入到A位置，如果整个区间左侧还有<A个位置可以插入数据，那么就只能将其放到整个区间的右侧，递归下去就可以了
1 void update(int k, int L, int R, int x) 2 { 3 if(L == R) { pre[k] = r; ans[L] = val; return ; } 4 5 int mid = (L+R)>>1; 6 7 if(x <= pre[k<<1]) update(lson, x);//左侧的多余x 8 9 else update(rson, x-pre[k<<1]);//否则往右 10 11 pre[k] = pre[k<<1] + pre[k<<1|1]; 12 }
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 1e9 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, mid 18 #define rson k<<1|1, mid+1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FOPENIN(IN) freopen(IN, "r", stdin) 23 #define FOPENOUT(OUT) freopen(OUT, "w", stdout) 24 template<class T> T CMP_MIN(T a, T b) { return a < b; } 25 template<class T> T CMP_MAX(T a, T b) { return a > b; } 26 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 27 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 28 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 29 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 30 31 typedef __int64 LL; 32 //typedef long long LL; 33 const int MAXN = 200005; 34 const int MAXM = 100005; 35 const double eps = 1e-10; 36 const LL MOD = 1000000007; 37 38 int N, pre[MAXN<<2]; 39 int id[MAXN], num[MAXN], ans[MAXN]; 40 int r, val; 41 42 int buildTree(int k, int L, int R) 43 { 44 if(L == R) return pre[k] = 1; 45 int mid = (L+R)>>1; 46 return pre[k] = buildTree(lson) + buildTree(rson); 47 } 48 49 void update(int k, int L, int R, int x) 50 { 51 if(L == R) { pre[k] = r; ans[L] = val; return ; } 52 53 int mid = (L+R)>>1; 54 55 if(x <= pre[k<<1]) update(lson, x); 56 57 else update(rson, x-pre[k<<1]); 58 59 pre[k] = pre[k<<1] + pre[k<<1|1]; 60 } 61 62 int main() 63 { 64 while(~scanf("%d", &N)) 65 { 66 buildTree(1, 1, N); 67 for(int l=1;l<=N;l++) 68 scanf("%d %d", &id[l], &num[l]); 69 r = 0; 70 for(int i=N;i>0;i--) 71 { 72 val = num[i]; 73 update(1, 1, N, id[i] + 1); 74 } 75 for(int i=1;i<=N;i++) printf("%d%c", ans[i], i==N?'\n':' '); 76 } 77 return 0; 78 }
