Uva 11991 - Easy Problem from Rujia Liu?(map 查找)
2013-11-27 00:19
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题目链接:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3142
题目分析:
每个询问都搜索一次,猜想超时过不了!于是将其弄成一一映射,用data[v][k]存第k个v的下标,询问的时候O(1)的输出,就不会超时喽!用STL中的map与vector来做
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3142
题目分析:
每个询问都搜索一次,猜想超时过不了!于是将其弄成一一映射,用data[v][k]存第k个v的下标,询问的时候O(1)的输出,就不会超时喽!用STL中的map与vector来做
#include<cstdio> #include<vector> #include<map> using namespace std; map<int, vector<int> > m; int main() { int n, t, temp, i, k, l; while(scanf("%d%d", &n, &t) == 2) { m.clear(); for(i = 0; i < n; i++) { scanf("%d", &temp); if (! m.count(temp)) m [temp] = vector<int>(); m [temp].push_back(i + 1); } while (t--) { scanf("%d%d", &k, &l); if(m[l].empty() || k > m[l].size()) printf("0\n"); else printf("%d\n", m[l][k-1]); } } return 0; }
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