UVA11991:Easy Problem from Rujia Liu?(STL构建变长二维数组)

11991 Easy Problem from Rujia Liu?
Though Rujia Liu usually sets hard problems for contests (for example, regional
contests like Xi’an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests like
Rujia Liu’s Presents 1 and 2), he occasionally sets easy problem (for example, ‘the
Coco-Cola Store’ in UVa OJ), to encourage more people to solve his problems :D
Given an array, your task is to find the k-th occurrence (from left to right) of an integer v. To make
the problem more difficult (and interesting!), you’ll have to answer m such queries.
Input
There are several test cases. The first line of each test case contains two integers n, m (1  n; m 
100; 000), the number of elements in the array, and the number of queries. The next line contains n
positive integers not larger than 1,000,000. Each of the following m lines contains two integer k and v
(1  k  n, 1  v  1; 000; 000). The input is terminated by end-of-file (EOF).
Output
For each query, print the 1-based location of the occurrence. If there is no such element, output ‘0’
instead.
Sample Input
8 4
1 3 2 2 4 3 2 1
1 3
2 4
3 2
4 2
Sample Output
2
0
7

0

思路很简单，但是直接开数组会超内存，一开始使用的vector也还是超内存

后来看大白书才发现要用map，map与vector结合成为一个变长数组

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 100005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8map<int,vector<int> > mat;int a;int main(){    int n,m,i,j,k,v;    while(~scanf("%d%d",&n,&m))    {        for(i = 1;i<=n;i++)        {            scanf("%d",&a);            if(!mat.count(a)) mat[a] = vector<int>();            mat[a].push_back(i);        }        while(m--)        {            scanf("%d%d",&k,&v);            if(!mat.count(v) || mat[v].size()<k) printf("0\n");            else printf("%d\n",mat[v][k-1]);        }    }    return 0;}