Gunner II

Gunner II"/>

Gunner II

题意描述

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.

给定n棵树和猎人能够进行的m次射击，每次射击能够射下相应高度最近树的鸟，询问落下鸟的顺序。

思路

我们可以使用结构体数组来存储树的序号与树的高度，使用两个数组来存储相应的序号和高度，同样可以使用lower_bound来寻找与射击高度相同的树，如果找到，则让对应的下标数组加1，没有找到则置-1。

下面来举个例子，对于样例

5 5
1 2 3 4 1
1 3 1 4 2

排序过后如图：

而对应的low数组和h数组如下：

low[1]=[1]，low[2]=5，low[3]=2，low[4]=3，low[5]=4
h[1]=1，h[2]=1，h[3]=2，h[4]=3，h[5]=4

如果寻找到第一颗高为1的树，则low[1]++成为2,当到第二个高为1的样例时，h[low[p]]=5，匹配成功。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long LL;
const int N=1e5+10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
struct node{int idx,h;
}Node[N];
bool cmp(node a,node b){if(a.h!=b.h) return a.h<b.h;return a.idx<b.idx;
}
int h[N],low[N];
void solve(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){for(int i=1;i<=n;i++){scanf("%d",&Node[i].h);Node[i].idx=i;}sort(Node+1,Node+1+n,cmp);for(int i=1;i<=n;i++) low[i]=i,h[i]=Node[i].h;for(int i=1;i<=m;i++){int x;scanf("%d",&x);int p=lower_bound(h+1,h+1+n,x)-h;if(h[low[p]]!=x) printf("-1\n");else{printf("%d\n",Node[low[p]].idx);low[p]++;}}}
}
int main(){//IOS;solve();return 0;
}