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 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>