P1135 奇怪的电梯 (双向bfs)

输入输出样例

输入 

5 1 5
3 3 1 2 5

输出

3

说明/提示

对于 100%100% 的数据，1≤N≤200，1≤A,B≤N，0≤Ki​≤N。

本题共 1616 个测试点，前 1515 个每个测试点 66 分，最后一个测试点 10 分。

1.重写AC代码：将步数记录在结构体中

#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

struct node{
	int u,d,step;
};
node a[210];

int cnt;
int vis[210];
int n,x,bb;
int path[210] = {-1};

bool bfs(int x){
	vis[x] = true;
	path[x] = 0;
	queue<node> q;
	q.push({a[x].u,a[x].d});
	
	while(!q.empty()){
		node temp;
		temp = q.front();
		q.pop();
		
		int up = temp.u;
		int down = temp.d;
		
		if(up != -1 && !vis[up]){
			vis[up] = true;
			a[up].step = temp.step + 1;
			q.push(a[up]);
		}
		if(down != -1 && !vis[down]){
			vis[down] = true;
			a[down].step = temp.step + 1;
			q.push(a[down]);
		}
		if(up == bb || down == bb){
			return true;
		}
	}
	return false;
}
int main()
{
	//要求从a到b 
	scanf("%d%d%d",&n,&x,&bb);
	for(int i=1;i<=n;i++){
		int c;
		scanf("%d",&c);
		a[i].u = c + i;
		a[i].d = i - c;

		if(a[i].u > n || a[i].u < 1) a[i].u = -1;
		if(a[i].d > n || a[i].d < 1) a[i].d = -1;
	} 
	if(bfs(x)) printf("%d",a[bb].step);
	else printf("-1");
	return 0;
}
 

结果：

2.AC代码：单设path数组记录到达每层楼用的步数

#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

struct node{
	int u,d,step,id;
};
node a[210];

int cnt;
int vis[210];
int n,x,bb;
int path[210] = {0};

bool bfs(int x){
	vis[x] = true;
	path[x] = 0;
	queue<node> q;
	q.push({a[x].u,a[x].d});
	
	while(!q.empty()){
		node temp;
		temp = q.front();
		q.pop();
		
		int up = temp.u;
		int down = temp.d;
		
		if(up != -1 && !vis[up]){
			vis[up] = true;
			//a[up].step = temp.step + 1;
			 path[up] = path[temp.id] + 1;
			q.push(a[up]);
		}
		if(down != -1 && !vis[down]){
			vis[down] = true;
			//a[down].step = temp.step + 1;
			path[down] = path[temp.id] + 1;
			q.push(a[down]);
		}
		if(up == bb || down == bb){
			return true;
		}
	}
	return false;
}
int main()
{
	//要求从a到b 
	scanf("%d%d%d",&n,&x,&bb);
	for(int i=1;i<=n;i++){
		int c;
		scanf("%d",&c);
		a[i].u = c + i;
		a[i].d = i - c;
		a[i].id = i;
		if(a[i].u > n || a[i].u < 1) a[i].u = -1;
		if(a[i].d > n || a[i].d < 1) a[i].d = -1;
	} 
	//if(bfs(x)) printf("%d",a[bb].step);
	if(bfs(x)) printf("%d",path[bb]);
	else printf("-1");
	return 0;
}
 

结果：