hdu 1548 A strange lift BFS 解法

 

 

//不用STL 写的BFS， 要用head 和 tail 标记 step 、 layer 数组。

// step 用来记录BFS 遍历的所有节点， layer 用来记录节点当前的层数 : layer[tail] = layer[head] + 1。

// 在一些要求记录最短路径的题型中，

// layer 可以用来记录前驱节点: layer[tail ] = head; //

// 然后 layer 从 后往前找到原路径，如 ： http://blog.youkuaiyun.com/hongxdong/archive/2010/03/19/5397192.aspx

1. 1. for ( int j = pre[tail]; j!= 0; j = pre[j])
   2. {
   3.         ans = ' ' + ans;
   4.         ans = char(way[j] + '0') + ans;
   5. }

 

#include <stdio.h>

using namespace std;


int main()
{
    int n, a, b;
    int step[1000];  // 记录节点列表.
    bool visited[1000];  // 判断节点是否访问过.
    int layer[10000];  // 记录每个节点所处搜索树的层数
    int dist[1000];  // 每层楼移动的距离.
    int head, tail, up, down;  // head 指向step 的头部, 用以访问保存在step 的节点.  tail 指向 step 的尾部用以添加节点.
    while (scanf("%d", &n) && n)
    {
        scanf ("%d %d", &a, &b);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &dist[i]);

            visited[i] = false;
        }
        if (a == b)
        {
            printf("0/n"); continue;
        }
        head = 0; tail = 1;
        step[head] = a;
        bool flag = false;
        while (head < tail)  // 节点访问未完成.
        {
            if (!visited[step[head]])  // step[head] 楼层未被访问
            {
                down = step[head] - dist[step[head]];  // 向下走.
                if (down > 0 && !visited[down])
                {
                    step[tail] = down;//保存未访问的节点

                    layer[tail] = layer[head] + 1;  // 保存上面节点的层数
                    if (step[tail] == b)  //如果到达b 楼层.
                    {
                        printf("%d/n", layer[tail]); // 打印出目标节点的层数, 即最少步数.
                        flag = true;
                        break;
                    }
                    tail++; // 后移step 尾部.
                }
                up = step[head] + dist[step[head]];
                if (up <= n && !visited[up])
                {
                    step[tail] = up;
                    layer[tail] = layer[head] + 1;
                    if (step[tail] == b)
                    {
                        printf("%d/n", layer[tail]);
                        flag = true;
                        break;
                    }
                    tail++;
                }
                visited[step[head]] = true;
            }
            head++; // 访问一个节点结束, 将step 的头部后移.
        }
        if (flag != true)
            printf("-1/n");
    }
}

下面是用STL 写的， 明显简单得多。

#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;

 

int main()
{
    int n, a, b;
    freopen("in.txt", "r", stdin);
    vector<int > v(1000);
    vector<bool > visited(1000);
    vector<int > step(1000);
    vector<int > f(10000);

    while (true)
    {
        cin>>n;
        if (n == 0)
            break;

        cin>>a>>b;

        for (int i = 1; i <= n; i++)
        {
            cin>>v[i];
            visited[i] = false;
            f[i] = 0;
        }
        if (a == b)
        {
            cout<<"0"<<endl;
            continue;
        }
        step.clear();
        step.push_back(a);

        int i = 0;
        bool flag = false;

        while (i != step.size())
        {
            if (!visited[step[i]])
            {
                if (step[i] == b)
                {
                    flag = true;
                    break;
                }
                if (step[i] - v[step[i]] > 0 && !visited[step[i] - v[step[i]]] )
                {
                    step.push_back(step[i] - v[step[i]]);
                    f[step.size() - 1] = f[i] + 1;
                }
                if (step[i] + v[step[i]] <= n && !visited[step[i] + v[step[i]]] )
                {
                    step.push_back(step[i] + v[step[i]]);
                    f[step.size() - 1] = f[i] + 1;
                }
                visited[step[i]] = true;
            }
            i++;
        }

        if (flag == true)
        {
            cout<<f[i]<<endl;
        }
        else
        {
            cout<<"-1"<<endl;
        }
    }
}