uva11991(二分查找或map的应用)

11991 - Easy Problem from Rujia Liu?

Time limit: 1.000 seconds

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). The size of input file does not exceed 5MB.

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

Output for the Sample Input

2
0
7
0

---

Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

简单题，我的方法是先带下标的排序，然后二分查找就行了。

这里用lower_bound()的时候需要用到它的第四个参数：

（这里有解释：http://msdn.microsoft.com/zh-cn/library/34hhk3zb.aspx）

comp

用户定义的谓词函数对象定义一个元素小于另一个。 二进制谓词采用两个参数，并且在满足时返回 true，在未满足时返回 false。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
#define INF 1000000000
#define eps 1e-8
#define pii pair<int,int>
#define LL long long int
struct node
{
    int id,val;
} a[];
int n,m,k,v;
bool cmp(node x,node y)
{
    if(x.val!=y.val)
        return x.val<y.val;
    else
        return x.id<y.id;
}
bool cmp2(node x,int y)
{
    return x.val<y;
}
int main()
{
    //freopen("in6.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&m)==)
    {
        for(int i=; i<n; i++)
        {
            scanf("%d",&a[i].val);
            a[i].id=i+;
        }
        sort(a,a+n,cmp);
        for(int i=;i<=m;i++)
        {
            scanf("%d%d",&k,&v);
            int t=lower_bound(a,a+n,v,cmp2)-a;
            if(t+k->=n||a[t+k-].val!=v)
            {
                printf("0\n");
            }
            else
            {
                printf("%d\n",a[t+k-].id);
            }
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return ;
}

大白书上提供了使用map的一种解法，很精彩的map应用，值得学习。

关键是要想清楚为什么这种思路需要用map：如果用数组来存，就必须开data[1000000][100000],开不了这么大。开数组时必须要考虑到每一维最大可能值然后开的不能小于这个值，其实有很多空间浪费了，因为这道题两维不会同时取到最大的。所以就用map在结合vector，用到什么就开什么，空间节约很多，而且map的查找和插入是logn的复杂度，时间可以保证。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
#define INF 1000000000
#define eps 1e-8
#define pii pair<int,int>
#define LL long long int
map<int,vector<int> >a;
int n,m,x,k,v;
int main()
{
    //freopen("in6.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&m)==)
    {
        a.clear();
        for(int i=;i<n;i++)
        {
            scanf("%d",&x);
            if(a.count(x)==) a[x]=vector<int>();
            a[x].push_back(i+);//a[x]就表示x的映射
        }
        for(int i=;i<m;i++)
        {
            scanf("%d%d",&k,&v);
            if(a.count(v)==||a[v].size()<k)    printf("0\n");
            else printf("%d\n",a[v][k-]);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return ;
}

注意：1.这里用的是a.count(x)来查找，这个函数就看二叉搜索树里有没有x，有返回1没有返回0.另一种查找方法是a.find(x)，如果有x，返回x所在的迭代器位置；没有则返回a.end().

2.数组法和insert()法在map里插入元素时对重复元素的处理不同。前者保持原来的，后者用新的覆盖原来的。

3.a[x]本身就是表示x的映射。