洛谷P1135 搜索

题目链接：传送门

一个水题，不过我dfs还是不太行，wa了两次
这题可以用DFS和BFS写，另外也可以建图跑最短路。下面放双搜代码：
BFS：

int a[210];
int A, B;
bool vis[210];
int n;
int ff = -1;

struct node
{
    int now, ans;
};
queue<node> q;
void bfs()
{
    while (q.size())
    {

        node temp = q.front();
        q.pop();
        vis[temp.now] = 1;
        if (temp.now == B)
        {
            ff = temp.ans;
            return;
        }
        if (temp.now + a[temp.now] <= n && vis[temp.now + a[temp.now]] == 0)
            q.push(node{temp.now + a[temp.now], temp.ans + 1});
        if (temp.now - a[temp.now] >= 1 && vis[temp.now - a[temp.now]] == 0)
            q.push(node{temp.now - a[temp.now], temp.ans + 1});
    }
}
int main()
{
    cin >> n >> A >> B;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    q.push(node{A, 0});
    bfs();
    cout << ff << endl;
    return 0;
}

DFS

int a[210];
int A, B;
bool vis[210];
int n;
int ff = INF;

void dfs(int now, int ans)
{
    if (ans > ff) //剪枝
        return;
    vis[now] = 1;
    if (now == B)
    {
        ff = min(ff, ans);
        vis[now] = 0; //回溯记得解标
        return;
    }
    int x = now + a[now];
    int y = now - a[now];
    if (x <= n && !vis[x])
        dfs(x, ans + 1);
    if (y >= 1 && !vis[y])
        dfs(y, ans + 1);
    vis[now] = 0;
}
int main()
{
    // fre;
    cin >> n >> A >> B;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    dfs(A, 0);
    if (ff == INF)
        cout << -1 << endl;
    else
        cout << ff << endl;
    return 0;
}