STL or 线段树 --- CSU 1555: Inversion Sequence
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Inversion SequenceProblem's Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1555
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Problem's Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1555
Mean:
给你一个序列a[n],要你按照要求去构造一个序列b。
序列a[i]表示序列b中的i前面有a[i]个数比i大。
转换一下就是:
已知一个连续的序列(1,2,3,4,...),然后告诉了我们这个序列中每个数前面比本身大的个数,根据这些条件将这个序列调整顺序,找到满足条件的序列。
analyse:
STL大法好。直接用vector来insert。
Time complexity: O(nlogn)
Source code:
// Memory Time// 1347K 0MS// by : crazyacking// 2015-03-29-23.24#include<map>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<cstdlib>#include<cstring>#include<climits>#include<iostream>#include<algorithm>#define MAXN 1000010#define LL long longusing namespace std;const int maxn = 11111; int a[maxn]; int ans[maxn]; vector<int>v; int main(){ // freopen("data.in","r",stdin); int n; while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++){ scanf("%d",&a[i]); } v.clear(); v.push_back(0); int ok=1; for(int i=n-1;i>=0;i--) { if(v.size()<=a[i]) { ok=0; break; } v.insert(v.begin()+1+a[i],i+1); // for(int j=0;j<v.size();j++) // printf("%d%c",v[j],j==v.size()-1?'\n':' '); } if(ok){ for(int i=1;i<=n;i++){ printf("%d%c",v[i],i==n?'\n':' '); } }else{ puts("No solution"); } } return 0; } /************************************************************** Problem: 1555 User: crazyacking Language: C++ Result: Accepted Time:164 ms Memory:1576 kb ****************************************************************/
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