uva--11991 - Easy Problem from Rujia Liu?（sort+二分 map+vector vector）

　　

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
题解：就是给你一个数组，问多次，每次找第k个v的坐标；我用了sort还用了二分，这竟然都没超时。。。数据太弱了吧。。。数据量是100000。。本来就是想试试 ，竟然过了，总结，以后别怕超时，别怂就是干。。。大神是map+vector，我用的vector  re了；；；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
#define F(i,s,x) for(i=s;i<x;i++)
const double PI=acos(-1.0);
typedef long long LL;
int n,m;
struct Node{
	int v,pos;
	friend bool operator < (Node a,Node b){
		if(a.v!=b.v)return a.v<b.v;
		else return a.pos<b.pos;
	}
}dt[100010];
int erfen(int x){
	int l=1,r=n,mid;
	while(l<=r){
		mid=(l+r)>>1;
		if(dt[mid].v>=x)r=mid-1;
		else l=mid+1;
	}
	return l;
}
int main(){
	while(~scanf("%d%d",&n,&m)){
		int i;
		F(i,1,n+1)SI(dt[i].v),dt[i].pos=i;
		sort(dt+1,dt+1+n);
		int k,v,pos;
	//	F(i,1,n+1)PI(dt[i].v),P_;puts("");
	//	F(i,1,n+1)PI(dt[i].pos),P_;puts("");
		while(m--){
			SI(k);SI(v);
			pos=erfen(v);
			//printf("%d\n",pos);
			if(dt[k+pos-1].v!=v)puts("0");
			else printf("%d\n",dt[k+pos-1].pos);
		}
	}
	return 0;
}

　　map+vetor;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
#define F(i,s,x) for(i=s;i<x;i++)
const double PI=acos(-1.0);
typedef long long LL;
const int MAXN=100010;
map<int,vector<int> >mp;
int main(){
	int m,n;
	while(~scanf("%d%d",&n,&m)){
		int i,x;
		F(i,1,n+1){
			SI(x);
		//	if(!mp.count(x))mp[x]=vector<int>();
			mp[x].push_back(i);
		}int k,v;
		while(m--){
			SI(k);SI(v);
			if(mp[v].size()<k)puts("0");
			else printf("%d\n",mp[v][k-1]);
		}
	}
	return 0;
}

　　vector  re...

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
#define F(i,s,x) for(i=s;i<x;i++)
const double PI=acos(-1.0);
typedef long long LL;
const int MAXN=100010;
vector<int>vec[MAXN];
int main(){
	int m,n;
	while(~scanf("%d%d",&n,&m)){
		int i,x;
		for(int i=0;i<MAXN;i++)vec[i].clear();
		F(i,1,n+1){
			SI(x);
			vec[x].push_back(i);
		}int k,v;
		while(m--){
			SI(k);SI(v);
			if(vec[v].size()<k)puts("0");
			else printf("%d\n",vec[v][k-1]);
		}
	}
	return 0;
}