HDU 1195 Open the Lock

Open the Lock

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1195
64-bit integer IO format: %I64d      Java class name: Main

 

 

Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. 
Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will change to be '1' and when minus 1 to '1', the digit will change to be '9'. You can also exchange the digit with its neighbor. Each action will take one step.

Now your task is to use minimal steps to open the lock.

Note: The leftmost digit is not the neighbor of the rightmost digit.

 

Input

The input file begins with an integer T, indicating the number of test cases. 

Each test case begins with a four digit N, indicating the initial state of the password lock. Then followed a line with anotther four dight M, indicating the password which can open the lock. There is one blank line after each test case.

 

Output

For each test case, print the minimal steps in one line.

 

Sample Input

2
1234
2144

1111
9999

Sample Output

2
4

Source

Zhejiang University Local Contest 2005

 

解题：bfs...

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
struct node{
    int key[4],step;
};
queue<node>q;
set<int>st;
int temp[4],tar[4];
bool transformIt(int *d){
    int num = 0;
    for(int i = 0; i < 4; i++)
        num = num*10 + d[i];
    if(st.count(num)) return false;
    st.insert(num);
    return true;
}
bool ok(int *d){
    for(int i = 0; i < 4; i++)
        if(d[i] != tar[i]) return false;
    return true;
}
int bfs(){
    while(!q.empty()) q.pop();
    node a,b;
    memcpy(a.key,temp,sizeof(temp));
    st.clear();
    a.step = 0;
    q.push(a);
    transformIt(a.key);
    while(!q.empty()){
        b = q.front();
        q.pop();
        if(ok(b.key)) return b.step;
        for(int i = 0; i < 4; i++){
            a.step = b.step+1;
            memcpy(temp,b.key,sizeof(temp));
            if(temp[i] + 1 > 9) temp[i] = 1;
            else temp[i]++;
            if(transformIt(temp)){
                memcpy(a.key,temp,sizeof(temp));
                q.push(a);
            }
            memcpy(temp,b.key,sizeof(temp));
            if(temp[i]-1 == 0) temp[i] = 9;
            else temp[i]--;
            if(transformIt(temp)){
                memcpy(a.key,temp,sizeof(temp));
                q.push(a);
            }
            if(i){
                memcpy(temp,b.key,sizeof(temp));
                swap(temp[i],temp[i-1]);
                if(transformIt(temp)){
                    memcpy(a.key,temp,sizeof(temp));
                    q.push(a);
                }
            }
            if(i < 3){
                memcpy(temp,b.key,sizeof(temp));
                swap(temp[i],temp[i+1]);
                if(transformIt(temp)){
                    memcpy(a.key,temp,sizeof(temp));
                    q.push(a);
                }
            }
        }
    }
    return -1;
}
int main(){
    int t,i,j;
    char md[6];
    scanf("%d",&t);
    while(t--){
        scanf("%s",md);
        for(i = 0; i < 4; i++) temp[i] = md[i] - '0';
        scanf("%s",md);
        for(i = 0; i < 4; i++) tar[i] = md[i] - '0';
        printf("%d\n",bfs());
    }
    return 0;
}

 

不用set居然快了这么多。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
struct node{
    int key[4],step;
};
queue<node>q;
int temp[4],tar[4];
bool vis[10000];
bool transformIt(int *d){
    int num = 0;
    for(int i = 0; i < 4; i++)
        num = num*10 + d[i];
    if(vis[num]) return false;
    vis[num] = true;
    return true;
}
bool ok(int *d){
    for(int i = 0; i < 4; i++)
        if(d[i] != tar[i]) return false;
    return true;
}
int bfs(){
    while(!q.empty()) q.pop();
    node a,b;
    memcpy(a.key,temp,sizeof(temp));
    memset(vis,false,sizeof(vis));
    a.step = 0;
    q.push(a);
    transformIt(a.key);
    while(!q.empty()){
        b = q.front();
        q.pop();
        if(ok(b.key)) return b.step;
        for(int i = 0; i < 4; i++){
            a.step = b.step+1;
            memcpy(temp,b.key,sizeof(temp));
            if(temp[i] + 1 > 9) temp[i] = 1;
            else temp[i]++;
            if(transformIt(temp)){
                memcpy(a.key,temp,sizeof(temp));
                q.push(a);
            }
            memcpy(temp,b.key,sizeof(temp));
            if(temp[i]-1 == 0) temp[i] = 9;
            else temp[i]--;
            if(transformIt(temp)){
                memcpy(a.key,temp,sizeof(temp));
                q.push(a);
            }
            if(i){
                memcpy(temp,b.key,sizeof(temp));
                swap(temp[i],temp[i-1]);
                if(transformIt(temp)){
                    memcpy(a.key,temp,sizeof(temp));
                    q.push(a);
                }
            }
            if(i < 3){
                memcpy(temp,b.key,sizeof(temp));
                swap(temp[i],temp[i+1]);
                if(transformIt(temp)){
                    memcpy(a.key,temp,sizeof(temp));
                    q.push(a);
                }
            }
        }
    }
    return -1;
}
int main(){
    int t,i,j;
    char md[6];
    scanf("%d",&t);
    while(t--){
        scanf("%s",md);
        for(i = 0; i < 4; i++) temp[i] = md[i] - '0';
        scanf("%s",md);
        for(i = 0; i < 4; i++) tar[i] = md[i] - '0';
        printf("%d\n",bfs());
    }
    return 0;
}