hdu 5233 Gunner II (map的简单用法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5233
Gunner II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1752 Accepted Submission(s): 639
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.
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)
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.
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.
Source
题目大意:上面一排数字,下面一排数字,找出下面一排数字在上面一排中的位置,如果有多个,输出优先靠左边的,然后删掉其数字。如果不存在输出-1
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <map> #include <vector> using namespace std; int h[100005], q[100005]; map<int,vector<int> > mp; int main () { int m, n; while(~scanf("%d %d", &n, &m)) { mp.clear(); for(int i=1; i<=n; i++) { scanf("%d", &h[i]); } for(int i=n; i>=1; i--) { mp[h[i]].push_back(i); } for(int i=1; i<=m; i++) { scanf("%d", &q[i]); if(mp[q[i]].size()>=1) { printf("%d\n",mp[q[i]].back()); mp[q[i]].pop_back(); } else printf("-1\n"); } } return 0; }