Uva 10085 The most distant state

Problem A The Most Distant State

Input: standard input Output: standard output

Time limit: 13.333 seconds

解题思路：题目的意思是给你一个初始化的3*3 “数字拼图”，叫你找出需要移动步数最多的那个拼图的情形，而且输出移动的步骤，当然移动步数最多的比较是建立在不同的拼图情形上的。题目为特判，所以输出可能有所不同。解题思路用BFS放队列里扫描“数字拼图”中空白处的上下左右方向，将拼图中数换回位数为9位的大整数，所以共有9！个整数，有出现的将其放到set中，并不断更新移动步骤的长度和当前的拼图情形，判断set中元素的个数是否达到9！break出来输出更新后的目标拼图情形和移动步骤

 

#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<set>
#define SIZE 9
#define INF 362880
#define MAXN 1000000
using namespace std;

typedef int subType[SIZE];
typedef struct type{
    subType value;
    string step;
}type;

set<int>visit;
queue<type>puzzle;

int dir[][2] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
string dirFlag[] = {"L", "R", "U", "D"};
int store_cnt;
int store_num;
string store_string;
bool try_to_insert(type from)
{
    int sum = 0;
    for(int i=0; i<SIZE; ++i) sum = sum*10 + from.value[i];
    if(visit.count(sum) == 0)
    {
        visit.insert(sum);
        if(store_string.length() <= from.step.length())
        {
            store_num = sum;
            store_string.assign(from.step);
        }
        store_cnt++;
        return true;
    }
    return false;
}

void Traverse()
{
    visit.clear();
    store_cnt = 0;
    store_string = "\0";
    store_num = 0;
    try_to_insert(puzzle.front());
    while(!puzzle.empty())
    {    
        type cur = puzzle.front();
        puzzle.pop();
        subType& s = cur.value;
        int z=0;
        for(; z<SIZE && s[z] != 0; ++z);
        int x=z/3, y = z%3;
        int flag = 0;
        for(int d=0; d<4; ++d)
        {
            int newx = x + dir[d][0];
            int newy = y + dir[d][1];
            int newz = newx * 3 + newy;
            if(newx>=0 && newx < 3 && newy >=0 && newy < 3)
            {
                type next;
                memcpy(next.value , s, sizeof(subType));
                next.value[newz] = s[z];
                next.value[z] = s[newz];
                next.step.assign(cur.step);
                next.step += dirFlag[d];
                if(try_to_insert(next))
                {
                    puzzle.push(next);
                    if(store_cnt == INF) 
                    {
                        flag = 1;
                        break;
                    }
                }
            }
        }
        if(flag) break;
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);

    #endif
    
    int T;
    cin>>T;
    
    for(int t=1; t<=T; ++t)
    {
        type input;
        for(int i=0; i<SIZE; ++i)
        {
            cin>>input.value[i];
        }
        input.step = "\0";
        while(!puzzle.empty()) puzzle.pop();
        puzzle.push(input);
        Traverse();
        cout<<"Puzzle #"<<t;
        for(int i=0, into=100000000; i<SIZE; ++i, into /= 10)
        {
            int temp = store_num/into;
            if(i%3 == 0) cout<<endl;
            else cout<<" ";
            cout<<temp;
            store_num %= into;
        }
        cout<<"\n"<<store_string<<endl;
        cout<<endl;
    }
    return 0;
}