【暑假】[实用数据结构]UVa11991 Easy Problem from Rujia Liu?
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UVa11991 Easy Problem from Rujia Liu?
思路: 构造数组data,使满足data[v][k]为第k个v的下标。因为不是每一个整数都会出现因此用到map,又因为每个数出现次数不等可能相差很大,因此用到vector。
注意:对于数据的清空与判空不要忘记,而map在调用之前必须有map.count的检查。
代码:
1 #include<cstdio> 2 #include<map> 3 #include<vector> 4 using namespace std; 5 6 int main(){ 7 int n,m; 8 map<int,vector<int> > table; 9 10 while(scanf("%d%d",&n,&m)==2) {11 table.clear();12 int x;13 for(int i=1;i<=n;i++) {14 scanf("%d",&x); //15 if(!table.count(x)) table[x] = vector<int>(); //map.count()的判断很重要 //vector的初始化 16 table[x].push_back(i);17 }18 int k,v; 19 for(int i=0;i<m;i++) {20 scanf("%d%d",&k,&v);21 if(!table.count(v) || table[v].size()<k) printf("0\n"); //在调用map之前定要count判断 22 else printf("%d\n",table[v][k-1]); //k-123 } 24 }25 return 0;26 }
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