poj 2828 Buy Tickets
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题目链接:http://poj.org/problem?id=2828
题目大意:给出每个人插入的位置,计算N个人插入队后,队中各个位置都是谁。
题目分析:用线段树动态的统计空余位置的数量和更新空余位置的信息。
代码参考:
#include<cmath>#include<vector>#include<string>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int M = 200009;int ans[M], pos[M], val[M];struct SegNode {int left, right, delay;void init(int a, int b){delay = b - a + 1;left = a;right = b;}} seg[M << 2];int L(int n){return n << 1;}int R(int n){return n << 1 | 1;}void build(int N, int a, int b){seg[N].init(a, b);if(a == b) { return ; }int mid = (a + b) >> 1;build(L(N), a, mid);build(R(N), mid + 1, b);}int update(int N, int id){seg[N].delay--;if(seg[N].left == seg[N].right) { return seg[N].left; }if(seg[L(N)].delay >= id) { update(L(N), id); }else { return update(R(N), id - seg[L(N)].delay); }}int main(){int n, sum, i, m;while(~scanf("%d", &n)) {for(i=0; i<n; ++i) {scanf("%d%d", &pos[i], &val[i]);}build(1, 1, n);for(i=n-1; i>=0; --i) {ans[update(1, pos[i] + 1)] = val[i];}for(i=1; i<=n; ++i) {printf("%d%c", ans[i], i == n ? '\n' : ' ');}}return 0;}
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