题目链接：http://icpc.njust.edu.cn/Problem/Hdu/1548/
题目大意：给定一个楼层、起始点和终点，每个楼可上可下一个长度，求重起始点到终点的最少上下次数。
代码如下：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef unsigned int ui; 4 typedef long long ll; 5 typedef unsigned long long ull; 6 #define pf printf 7 #define mem(a,b) memset(a,b,sizeof(a)) 8 #define prime1 1e9+7 9 #define prime2 1e9+9 10 #define pi 3.14159265 11 #define lson l,mid,rt<<1 12 #define rson mid+1,r,rt<<1|1 13 #define scand(x) scanf("%llf",&x) 14 #define f(i,a,b) for(int i=a;i<=b;i++) 15 #define scan(a) scanf("%d",&a) 16 #define dbg(args) cout<<#args<<":"<<args<<endl; 17 #define inf 0x3f3f3f3f 18 #define maxn 1005 19 #define maxm 2000010 20 int n,m,t,a,b; 21 int st,ed; 22 int k[maxn]; 23 bool vis[maxn]; 24 struct node{ 25 int x,step; 26 node(int x,int step):x(x),step(step){}; 27 node(){}; 28 }; 29 node cur,nxt; 30 int bfs() 31 { 32 queue<node>q; 33 node st(a,0); 34 vis[a]=true;//一旦到达该楼层 35 q.push(st); 36 while(!q.empty()) 37 { 38 cur=q.front(); 39 q.pop(); 40 if(cur.x==b) 41 { 42 return cur.step; 43 //由于队列前面的状态结点的step值一定大于等于后面结点的step值，所以第一次达到目标状态的step一定是最优值 44 } 45 if(cur.x+k[cur.x]<=n&&cur.x+k[cur.x]>0&&!vis[cur.x+k[cur.x]]) 46 { 47 vis[cur.x+k[cur.x]]=true; 48 nxt.x=cur.x+k[cur.x]; 49 nxt.step=cur.step+1; 50 q.push(nxt); 51 } 52 if(cur.x-k[cur.x]<=n&&cur.x-k[cur.x]>0&&!vis[cur.x-k[cur.x]]) 53 { 54 vis[cur.x-k[cur.x]]=true; 55 nxt.x=cur.x-k[cur.x]; 56 nxt.step=cur.step+1; 57 q.push(nxt); 58 } 59 } 60 return -1; 61 } 62 int main() 63 { 64 //freopen("input.txt","r",stdin); 65 //freopen("output.txt","w",stdout); 66 std::ios::sync_with_stdio(false); 67 while(scan(n)&&n) 68 { 69 scan(a); 70 scan(b); 71 mem(vis,false); 72 f(i,1,n)scan(k[i]); 73 pf("%d ",bfs()); 74 } 75 }