POJ 2828Buy Tickets

POJ 2828

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

 void update(int k, int L, int R, int x)
 {
     if(L == R) { pre[k] = r; ans[L] = val; return ; }

     int mid = (L+R)>>;

     if(x <= pre[k<<]) update(lson, x);//左侧的多余x

     else update(rson, x-pre[k<<]);//否则往右

      pre[k] = pre[k<<] + pre[k<<|];
 }

 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <vector>
 #include <cstdio>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define INF 1e9
 #define inf (-((LL)1<<40))
 #define lson k<<1, L, mid
 #define rson k<<1|1, mid+1, R
 #define mem0(a) memset(a,0,sizeof(a))
 #define mem1(a) memset(a,-1,sizeof(a))
 #define mem(a, b) memset(a, b, sizeof(a))
 #define FOPENIN(IN) freopen(IN, "r", stdin)
 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

 typedef __int64 LL;
 //typedef long long LL;
 const int MAXN = ;
 const int MAXM = ;
 const double eps = 1e-;
 const LL MOD = ;

 int N, pre[MAXN<<];
 int id[MAXN], num[MAXN], ans[MAXN];
 int r, val;

 int buildTree(int k, int L, int R)
 {
     if(L == R) return pre[k] = ;
     int mid = (L+R)>>;
     return pre[k] = buildTree(lson) + buildTree(rson);
 }

 void update(int k, int L, int R, int x)
 {
     if(L == R) { pre[k] = r; ans[L] = val; return ; }

     int mid = (L+R)>>;

     if(x <= pre[k<<]) update(lson, x);

     else update(rson, x-pre[k<<]);

      pre[k] = pre[k<<] + pre[k<<|];
 }

 int main()
 {
     while(~scanf("%d", &N))
     {
         buildTree(, , N);
         for(int l=;l<=N;l++)
             scanf("%d %d", &id[l], &num[l]);
         r = ;
         for(int i=N;i>;i--)
         {
             val = num[i];
             update(, , N, id[i] + );
         }
         for(int i=;i<=N;i++) printf("%d%c", ans[i], i==N?'\n':' ');
     }
     return ;
 }