标签:ret idt turn ack isp 位置 tick line div
题目链接:http://poj.org/problem?id=2828
插队买票,有 n 个人依次插队,一开始买票队列为空,每个插队的人有两个属性 (pos, val) ,pos 表示这个人插队插在了 pos 位置,val 代表这个人的 id,输出 n 个人插完队后形成的序列所对应的 id 序列。
对于 0 31492,应该填在 0 号位置:
然后把 31492 抽掉:
接着放 1 3890,应该填在 1 号位置:
然后把 3890抽掉:
接着放 1 19243,应该填在 1 号位置:
接着放 0 20523,应该填在 0 号位置:
抽掉后列表为空,表明插完了,总结果如下:
线段树每个节点维护 1,如果该位被抽掉了,就替换为 0。
1 #pragma GCC optimize("Ofast") 2 #include <bits/stdc++.h> 3 using namespace std; 4 5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 6 #define Rep(i,n) for (int i = 0; i < (n); ++i) 7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 13 14 #define pr(x) cout << #x << " = " << x << " " 15 #define prln(x) cout << #x << " = " << x << endl 16 17 #define LOWBIT(x) ((x)&(-x)) 18 19 #define ALL(x) x.begin(),x.end() 20 #define INS(x) inserter(x,x.begin()) 21 22 #define ms0(a) memset(a,0,sizeof(a)) 23 #define msI(a) memset(a,inf,sizeof(a)) 24 #define msM(a) memset(a,-1,sizeof(a)) 25 26 #define MP make_pair 27 #define PB push_back 28 #define ft first 29 #define sd second 30 31 template<typename T1, typename T2> 32 istream &operator>>(istream &in, pair<T1, T2> &p) { 33 in >> p.first >> p.second; 34 return in; 35 } 36 37 template<typename T> 38 istream &operator>>(istream &in, vector<T> &v) { 39 for (auto &x: v) 40 in >> x; 41 return in; 42 } 43 44 template<typename T1, typename T2> 45 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 46 out << "[" << p.first << ", " << p.second << "]" << "\n"; 47 return out; 48 } 49 50 typedef long long LL; 51 typedef unsigned long long uLL; 52 typedef pair< double, double > PDD; 53 typedef pair< int, int > PII; 54 typedef set< int > SI; 55 typedef vector< int > VI; 56 typedef map< int, int > MII; 57 typedef vector< LL > VL; 58 typedef vector< VL > VVL; 59 const double EPS = 1e-10; 60 const int inf = 1e9 + 9; 61 const LL mod = 1e9 + 7; 62 const int maxN = 2e5 + 7; 63 const LL ONE = 1; 64 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 65 const LL oddBits = 0x5555555555555555; 66 67 int n; 68 int pos[maxN], val[maxN], ans[maxN]; 69 70 #define lson l , mid , rt << 1 71 #define rson mid + 1 , r , rt << 1 | 1 72 73 struct SegmentTree{ 74 int st[maxN << 2]; 75 76 inline void pushUp(int rt) { 77 st[rt] = st[rt << 1] + st[rt << 1 | 1]; 78 } 79 80 inline void pushDown(int rt) { } 81 82 inline void build(int l, int r, int rt) { 83 if(l >= r) { 84 st[rt] = 1; 85 return; 86 } 87 int mid = (l + r) >> 1; 88 build(lson); 89 build(rson); 90 pushUp(rt); 91 } 92 93 // 把相应位置设为 0 94 inline void update(int x, int l, int r, int rt) { 95 if(l >= r) { 96 st[rt] = 0; 97 return; 98 } 99 int mid = (l + r) >> 1; 100 if(x <= mid) update(x, lson); 101 else update(x, rson); 102 pushUp(rt); 103 } 104 105 // 查找从 1 开始和为 x 的区间 [1, r] 的右端点 r 106 inline int querySum(int x, int l, int r, int rt) { 107 if(l >= r) return r; 108 int mid = (l + r) >> 1; 109 if(st[rt << 1] >= x) return querySum(x, lson); 110 else return querySum(x - st[rt << 1], rson); 111 } 112 }; 113 SegmentTree segTr; 114 115 int main(){ 116 INIT(); 117 while(cin >> n) { 118 For(i, 1, n) { 119 cin >> pos[i] >> val[i]; 120 ++pos[i]; 121 } 122 segTr.build(1, n, 1); 123 124 rFor(i, n, 1) { 125 int tmp = segTr.querySum(pos[i], 1, n, 1); 126 ans[tmp] = val[i]; 127 segTr.update(tmp, 1, n, 1); 128 } 129 130 For(i, 1, n) cout << ans[i] << " "; 131 cout << endl; 132 } 133 return 0; 134 }
标签:ret idt turn ack isp 位置 tick line div
原文地址:https://www.cnblogs.com/zaq19970105/p/10798298.html