hdu5233 邻接表 两组代码，一组AC，一组不知道为什么RE了（以后再看）

Gunner II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1622    Accepted Submission(s): 607

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 51 2 3 4 11 3 1 4 2

 

Sample Output

13542
Hint
Huge input, fast IO is recommended.
 

题目大意：

       有n棵排成一排树，有n只鸟分别在编号对于的树上。这些树都有具体的树高。一个猎人打猎，站在树的排头，每次打出的高度是q米，问能打下来那棵树上的鸟。

遇到的问题和思路：

       刚开始打算用vector的二维数组来做。。。结果发现只能呵呵了，表示清空不会，然后vector的做法莫名其妙的经常RE，感觉很奇怪。

      于是看了一下别人的做法，map中有vector，因为map.clear（）很轻松就能清空了，感觉很方便。

#include<cstdio>#include<algorithm>#include<map>#include<queue>using namespace std;int n, m, i_max;map <int, queue<int> > hq;int q[100000 + 10];void solve(){for (int i = 1; i <= m; i++){if(q[i] > i_max || hq[q[i]].empty()){printf("-1\n");continue;}printf("%d\n", hq[q[i]].front());hq[q[i]].pop();}}int main(){while(scanf("%d %d", &n, &m) != EOF){memset(q, 0, sizeof(q));hq.clear();i_max = 0;for (int i = 1; i <= n; i++){int H;scanf("%d", &H);hq[H].push(i);i_max = max(i_max, H);}for (int i = 1; i <= m; i++){scanf("%d", q + i);}solve();}return 0;}//不知道为什么，下面这个一直RE，铭铭姐说清空这里出现了问题，但是我把清空这里/**/了也还是RE（虽然我知道不清空肯定是错的）。感觉有点不解，等我以后成了大牛再回来看吧（小小的期待，不过要付出努力） ^0^//给出RE的代码：#include<cstdio>#include<algorithm>#include<vector>#include<cstring>using namespace std;int n, m, i_max;vector <int> h[100000 + 300];int q[100000 + 300];void solve(){for (int i = 0; i < m; i++){if (h[q[i]].size() == 0 || q[i] > i_max){//比最高的还要高，或者是这个树高的树已经没了 printf("-1\n");continue;}printf("%d\n", h[q[i]][0]);h[q[i]].erase(h[q[i]].begin());}/*for (int i = 0; i < n; i++){//清除元素while (h[q[i]].size()){h[q[i]].pop_back();}} */}int main(){while(scanf("%d %d", &n, &m) != EOF){memset(q, 0, sizeof(q));i_max = 0;for (int i = 0; i < n; i++){int H;scanf("%d", &H);i_max = max(i_max, H);h[H].push_back(i + 1);}for (int i = 0; i < m; i++){scanf("%d", &q[i]);}solve();}return 0;}</int></cstring></vector></algorithm></cstdio></int></queue></map></algorithm></cstdio>