UVA 10085 The most distant state

 

大意：八数码问题的变形，让你求离当前状态最远的距离。

思路：BFS + hash判重，直到不能扩展为止，最后一个节点一定是最远的距离（由BFS性质知道）

CODE：

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
using namespace std;


typedef int State[9];
const int MAXN = 1000003;
const int dx[] = {-1,1,0,0};
const int dy[] = {0,0,-1,1};
char dir[5] = "UDLR";

State st[MAXN];
int fa[MAXN], path[MAXN];     //存储路径 
int first[MAXN], next[MAXN];
State state;

int ans;

void init()
{
    memset(first, -1, sizeof(first));
    memset(fa, 0, sizeof(fa));
    memset(path, 0, sizeof(path));
}

int hash(State &s) //映射 
{
    int v = 0;
    for(int i = 0; i < 9; i++) v = v*10 + s[i];
    return v%MAXN;
}

int try_to_insert(int s)   //hash判重 
{
    int h = hash(st[s]);
    for(int v = first[h]; v!=-1; v = next[v])
    {
        if(memcmp(st[v], st[s], sizeof(st[s])) == 0) return 0;
    }
    next[s] = first[h];
    first[h] = s;
    return 1;
}

int check(int r, int c)
{
    if(r >= 0 && r < 3 && c >= 0 && c < 3)    return 1;
     return 0;
}

void bfs()
{
    init();
    int front = 0, rear = 1;
    fa[0] = path[0] = -1;
    try_to_insert(0);
    while(front < rear)
    {
        State& s = st[front];
        int z;
        for(z = 0; z < 9; z++) if(!s[z]) break;
        int x = z/3, y = z%3;
        for(int i = 0; i < 4; i++)
        {
            int newx = x + dx[i];
            int newy = y + dy[i];
            int newz = 3*newx + newy;
            if(check(newx, newy))
            {
                State& t = st[rear];
                memcpy(&t, &s, sizeof(s));
                t[newz] = s[z];
                t[z] = s[newz];
                if(try_to_insert(rear))
                {
                    fa[rear] = front;
                    path[rear] = i;
                    rear++;
                }
            }
        }
        front++;
    }
    ans = rear-1;
}

void print_path(int cur) //递归打印路径 
{
    if(cur)
    {
        print_path(fa[cur]);
        printf("%c", dir[path[cur]]);
    }
}

int main()
{
    int T, times = 0;
    scanf("%d", &T);
    while(T--)
    {
        for(int i = 0; i < 9; i++)    scanf("%d", &st[0][i]);
        bfs();
        printf("Puzzle #%d\n", ++times);
        for(int i = 0; i < 3; i++)
        {
            printf("%d %d %d", st[ans][3*i], st[ans][3*i+1], st[ans][3*i+2]);
            printf("\n");
        }
        print_path(ans);
        printf("\n\n");
    }
    return 0;
}