題干：

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.?

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute?
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.?

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

解題報告

? ?這題用bfs去遞推（其實也相當于dp啦，可以理解成? 只是不是按照for循環遍歷的順序去dp而已）

AC代碼：

#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #define ll long long using namespace std; const int INF = 0x3f3f3f3f; bool vis[300000 + 5]; int n,k; struct Node {int pos,step;Node(){}Node(int pos,int step):pos(pos),step(step){} }; int bfs(int x) {queue<Node> q;memset(vis,0,sizeof vis);q.push(Node(x,0));vis[x]=1;while(!q.empty()) {Node cur = q.front();q.pop();if(cur.pos == k) return cur.step;if( cur.pos+1<= 300000 && cur.pos+1 >= 0&&!vis[cur.pos+1] ){vis[cur.pos+1]=1;q.push(Node(cur.pos+1,cur.step+1));}if(cur.pos-1<= 300000&& cur.pos-1 >= 0 &&!vis[cur.pos-1] ) {vis[cur.pos-1]=1;q.push(Node(cur.pos-1,cur.step+1));}if((cur.pos<<1) <= 300000 && (cur.pos<<1) >=0 &&!vis[cur.pos<<1] ) {vis[cur.pos<<1]=1 ;q.push(Node(cur.pos<<1,cur.step+1)); }}return 0; } int main() {while(~scanf("%d%d",&n,&k)) {printf("%d\n",bfs(n));}return 0; }

錯誤代碼：（dp）

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define ll long long using namespace std; const int INF = 0x3f3f3f3f; ll dp[3000000 +5]; inline ll min3(ll i,ll j,ll k) {if(i < j) {return i < k ? i : k;}if(j < k) return j;else return k; } int main() {int n,k;while(~scanf("%d%d",&n,&k)) {memset(dp,INF,sizeof dp);for(int i = 1; i<=n; i++) {dp[i] = min(dp[i],(ll)n-(ll)i);dp[i*2] = min(dp[i*2],dp[i]+1);}if(k <= n) {printf("%lld\n",dp[k]);continue;}for(int i = n+1; i<=(k<<1); i++) {dp[i] = min3(dp[i],dp[i-1]+ 1,dp[i+1] + 1) ;dp[i*2] = min(dp[i*2],dp[i]+1);}printf("%lld\n",dp[k]);}return 0; }

總結：

? ?這題有個很大的坑啊！！我剛開始不管數組開多大都是re，后來發現，，，數組不需要開多大，因為他有可能是負數！！所以需要判斷是否越界。

? ?其二，開30W的數組就夠了，不需要300W，但是有一點要注意，判斷條件中，必須判斷范圍在前，判斷vis數組在后。

? ? ? 即，判斷越界在前，判斷其他在后。這一點非常重要！我立馬想到了一般我寫地圖類dfs中的tx，ty也沒大注意先判斷越界還是先判斷maze是否滿足還是先判斷vis是否標記過，這是因為！！！我的地圖都是從1開始讀入的，所以就算是先判斷別的在判斷txty是否越界，也不會出現re的情況，這一點是我今天才注意到的、、慚愧慚愧、、、if中判斷順序有的時候也是有區別的！！！有可能會導致越界！！

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