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
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 using namespace std;
7 const int INF=0x3f3f3f3f;
8 #define SI(x) scanf("%d",&x)
9 #define PI(x) printf("%d",x)
10 #define P_ printf(" ")
11 #define mem(x,y) memset(x,y,sizeof(x))
12 const int MAXN=5000010;
13 int tp;
14 int hsh[MAXN],head[MAXN],pos[MAXN],nxt[MAXN];
15 int ans[100010];
16 int q[100010];
17 void add(int p,int x){
18 int i=x%MAXN;
19 hsh[tp]=x;
20 pos[tp]=p;
21 nxt[tp]=head[i];
22 head[i]=tp++;
23 }
24 int main(){
25 int n,m;
26 while(~scanf("%d%d",&n,&m)){
27 int x;
28 mem(head,-1);tp=0;
29 mem(nxt,0);mem(hsh,0);
30 for(int i=1;i<=n;i++){
31 scanf("%d",&x);add(i,x);
32 }
33 for(int i=1;i<=m;i++)scanf("%d",&q[i]);
34 for(int i=m;i>=1;i--){
35 x=q[i];
36 int flot=0;
37 for(int j=head[x%MAXN];j!=-1;j=nxt[j]){
38 if(x==hsh[j]){
39 ans[i]=pos[j];
40 flot=1;
41 hsh[j]=0;break;
42 }
43 }
44 if(!flot)ans[i]=-1;
45 }
46 for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
47 }
48 return 0;
49 }