uva10085

来源：互联网发布：深圳网络诈骗举报编辑：程序博客网时间：2024/06/10 11:52

题目大意：
这一题跟八编码的题目有些类似，但是它并不是求目标状态，它所求的是可以达到的最远的那个图。

思路：

第一次接触这样的题目，看了题解才过的。
使用到了哈希判重和bfs，还需要递归的打印路径。

#include <iostream>using namespace std;#include <cstring>#include <stdio.h>typedef int State[9];const int MAXSTATE = 1000000;int state[9];State st[MAXSTATE],goal;int father[MAXSTATE],path[MAXSTATE],ans;char path_dir[5] ="UDLR";int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};const int MAXHASHSIZE = 1000003;int head[MAXHASHSIZE],next1[MAXSTATE];void init_lookup_table () {    memset(head,0,sizeof(head));}int hash1(State &s );int try_to_insert(int s) {    int h = hash1(st[s]);    int u = head[h];    while(u) {        if(memcmp(st[u],st[s],sizeof(st[s])) == 0) return 0;        u = next1[u];    }    next1[s] = head[h];    head[h] = s;    return 1;}void bfs() {    init_lookup_table();    father[0] = path[0] =-1;    int front = 0 ,rear =1;    memcpy(st[0],state,sizeof(state));    try_to_insert(0);    while(front < rear) {        int *s =st[front];        int j;        for(j = 0; j <9; j++)            if(!s[j])                break;        int x = j/3,y = j%3;        for(int i = 0 ; i <4; i++) {            int dx = x+ dir[i][0];            int dy = y + dir[i][1];            int pos = dx * 3 + dy;            if(dx >= 0 && dx <3 && dy >=0 && dy <3) {                int *newState = st[rear];                memcpy(newState,s,sizeof(int)*9);                newState[j] = s[pos];                newState[pos] = 0;                if(try_to_insert(rear)) {                    father[rear] = front;                    path[rear] = i;                    rear++;                }            }        }        front ++;    }    ans = rear -1;}int hash1(State &s) {    int v = 0;    for(int i = 0; i <9 ;i++)        v = v * 10 + s[i];    return v% MAXHASHSIZE;}void print_path(int cur) {    if(cur!= 0) {        print_path(father[cur]);        printf("%c",path_dir[path[cur]]);    }}int main() {    int T,cas = 1;    scanf("%d",&T);    while(T--) {        for(int i = 0; i < 9; i++)            scanf("%d",&state[i]);        bfs();        printf("Puzzle #%d\n",cas++);        for(int i = 0; i <3; i++)            printf("%d %d %d\n",st[ans][i*3],st[ans][i*3+1],st[ans][i*3+2]);        print_path(ans);        printf("\n\n");    }    return 0;}

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