动态规划训练13 [Catch That Cow poj3278]

Catch That Cow

POJ - 3278

这道题我看大家用的方法都是bfs搜索，为什么在我看来这就是一个动态规划的题目啊啊啊啊啊啊啊

dp[x]表示从N出发到x所需要的最小时间

那么得到如下转移方程

如果x < N的话，那么只能通过走路来转移，所以dp[x] = N - x，x <= N时候

而x > N时候，可以通过2种方式来转移

（1）走路转移 dp[x] = dp[x-1] + 1

（2）跳跃加走路转移

当x为偶数的时候

dp[x] = min(dp[x],dp[x/2]+1,dp[x/2+1]+3)

当x为奇数的时候

dp[x] = min(dp[x],dp[(x-1)/2]+2,dp[(x+1)/2]+2)

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int N,K;
const int MAX = 100005;
int dp[MAX];int main(){cin>>N>>K;for(int i = 0;i <= K;i++){dp[i] = abs((int)(N - i));}for(int i = N+1;i <= K;i++){if(i & 1) // 奇数{dp[i] = min(dp[i],dp[(i-1)/2]+2);dp[i] = min(dp[i],dp[i-1]+1);dp[i] = min(dp[i],dp[(i+1)/2]+2);} else{dp[i] = min(dp[i],dp[i/2]+1);dp[i] = min(dp[i],dp[i-1]+1);dp[i] = min(dp[i],dp[i/2 + 1]+3);}}cout<<dp[K]<<endl;return 0;
}

补充：刚才尝试了一下模拟的方法，确实也行的通，而且代码量不大

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
typedef pair<int,int> P;
const int MAX = 200005;
int used[MAX];
int main(){
    int N,K;
    cin>>N>>K;
    queue<P> Q;
    Q.push(make_pair(0,N));
    while(!Q.empty()){
        P p = Q.front();Q.pop();
        if(p.second == K){
            cout<<p.first<<endl;
            break;
        }
        if(p.second <= K + 2  && !used[p.second + 1]){
            Q.push(make_pair(p.first+1,p.second + 1));
            used[p.second + 1] = 1;
        }
        if(p.second >= 1 && !used[p.second - 1]){
            Q.push(make_pair(p.first+1,p.second - 1));
            used[p.second - 1] = 1;
        }
        if(p.second * 2 <= 2* K  && !used[p.second * 2]){
            Q.push(make_pair(p.first+1,p.second * 2));
            used[p.second * 2] = 1;
        }
    }
    return 0;
}

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动态规划训练13 [Catch That Cow poj3278]

Catch That Cow

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