Gunner II（二分，map,数字转化）

Gunner II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1740    Accepted Submission(s): 635

Problem Description

Long long ago, there was a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i-th bird stands on the top of the i-th tree. The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the nearest bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know which bird will fall during each shot.

 

 

Input

There are multiple test cases (about 5), every case gives n, m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times. 
In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.
In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.
Please process to the end of file.
[Technical Specification]
All input items are integers.
1<=n,m<=100000(10^5)
1<=h[i],q[i]<=1000000000(10^9)

 

 

Output

For each q[i], output an integer in a single line indicates the id of bird Jack shots down. If Jack can’t shot any bird, just output -1.
The id starts from 1.

 

 

Sample Input

5 5 1 2 3 4 1 1 3 1 4 2

 

 

Sample Output

1 3 5 4 2

Hint

Huge input, fast IO is recommended.

 

题解：这题就做的曲折了，刚开始一看不就是个hash表么，倒着记录下就好了，然而在tel和mel中，我放弃了。。。

于是就想着二分了，对于已经用过的怎么办，再开个数组记录已经用过的，可是数组太大了啊，于是我用map来记录，可是是数字啊，木事木事，转化成string不就好了。。。想用stl里的二分类，但是是结构体啊，那就自己写。。。

写完了，感觉很有可能wa，因为二分感觉写的有点挫，交了下竟然AC了。。。我只想笑。。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
struct Node{
    int v,num;
    bool operator < (const Node &b) const{
        if(v!=b.v){
            return v<b.v;
        }
        else return num<b.num;
    }
};
Node dt[MAXN];

int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        map<string,int>mp;
        for(int i=1;i<=n;i++)SI(dt[i].v),dt[i].num=i;
        sort(dt+1,dt+n+1);
        int x;
        sort(dt+1,dt+n+1);
        for(int i=0;i<m;i++){
            SI(x);
            char s[10];
            itoa(x,s,10);
            int l=1,r=n+1,mid;
            while(l<=r){
                mid=(l+r)>>1;
                if(x<=dt[mid].v)r=mid-1;
                else l=mid+1;
            }
            if(dt[r+1+mp[s]].v==x){
                printf("%d\n",dt[r+1+mp[s]].num);
                mp[s]++;
            }
            else puts("-1");
        }
    }
    return 0;
}

我的hash表，TLE了：（贡献了9次TLE，MLE）贴下吧

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=5000010;
int tp;
int hsh[MAXN],head[MAXN],pos[MAXN],nxt[MAXN];
int ans[100010];
int q[100010];
void add(int p,int x){
    int i=x%MAXN;
    hsh[tp]=x;
    pos[tp]=p;
    nxt[tp]=head[i];
    head[i]=tp++;
}
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        int x;
        mem(head,-1);tp=0;
        mem(nxt,0);mem(hsh,0);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);add(i,x);
        }
        for(int i=1;i<=m;i++)scanf("%d",&q[i]);
        for(int i=m;i>=1;i--){
                x=q[i];
            int flot=0;
            for(int j=head[x%MAXN];j!=-1;j=nxt[j]){
                if(x==hsh[j]){
                    ans[i]=pos[j];
                    flot=1;
                    hsh[j]=0;break;
                }
            }
            if(!flot)ans[i]=-1;
        }
        for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
    }
    return 0;
}