HDU 5233 Gunner II 数据结构map+vector

B - Gunner II

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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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 51 2 3 4 11 3 1 4 2 

 

Sample Output

13542 

Hint

Huge input, fast IO is recommended. 

解题思路：

做这题的时候错了很多次，这题主要就是：

1，鸟的高度太多了，需要map进行映射，然后存储每一个鸟的位置。

2，同一个高度可能有很多的鸟，打下来的是下标最小的，所以需要一个链式的结构去存，不能用二维数组

因为二维数组开不了那么大，所以就用vector了。

3，对于每一个vector打掉一只鸟不能用erase删除元素，那么就不删除元素，再定义一个数组，记录每一个vector的最左边的鸟的位置即可。

#include<cstdio>#include<cstring>#include<iostream>#include<map>#include<vector>using namespace std;const int maxn = 100005 ;int h[maxn] ;int q[maxn] ;int x[maxn] ;vector<int> s[maxn] ;map<int,int>qq ;int main(){    int n,m;    while(~scanf("%d%d",&n,&m)){        qq.clear();        int c = 1 ;        for(int i=0;i<n;i++){            scanf("%d",&h[i]);            if(!qq.count(h[i])){                qq[h[i]] = c++;                s[qq[h[i]]].clear();                x[qq[h[i]]] = 0 ;            }            s[qq[h[i]]].push_back(i+1);        }        for(int i=0;i<m;i++){            scanf("%d",&q[i]);            //printf("as = %d %d\n",qq[q[i]],qq.count(q[i]));            if(x[qq[q[i]]]>=s[qq[q[i]]].size()||!qq.count(q[i]))printf("-1\n");            else{                printf("%d\n",s[qq[q[i]]][x[qq[q[i]]]]);                x[qq[q[i]]]++ ;            }        }    }    return 0;}