UVa1601 Morning after holloween 单向BFS方法

　　紫书上的题，一开始全用stl容器结果tle，于是重写一遍全换成自己手写的容器。另外重写判重和互穿的时候还将n=1,2,3时的情况单独分类。AC代码如下。这道题给的内存还是很充足的。

#include<iostream>
#include<memory.h>
#include<string>
#define INIT(x) memset(x,0,sizeof(x))
using namespace std;

int w, h, n;
int num[20][20];//每个空格的编号
int conn[200][200];//邻接表
int visited[200][200][200];
char G[20][20];
int goal[3];
int start[3];
int queue[10000000][4],front,rear;
void queIns(int a[3],int step)
{
    queue[rear][0] = step;
    queue[rear][1] = a[0];
    queue[rear][2] = a[1];
    queue[rear][3] = a[2];
    rear += 1;
}
void quePop(){++front;}
int* getQueFront(){
    return queue[front];
}
bool queIsEmpty(){return front == rear;}

void init()
{
    INIT(num);
    INIT(conn);
    INIT(visited);
    INIT(G);
    INIT(goal);
    INIT(start);
    INIT(queue);
    front = 0; rear = 0;
}

bool isOverlap(int a[3])//是否位置重叠
{
    if (n == 3)
    {
        return (a[1] == a[2] || a[1] == a[0] || a[2] == a[0]);
    }
    else if (n == 2)
    {
        return (a[1] == a[0]);
    }
    else if (n == 1)
    {
        return false;
    }
}

bool isAvalible(int a[3], int b[3])//移动是否可行（防止互穿）
{
    if (n == 3)
    {
        if (a[1] == b[2] && a[2] == b[1])return false;
        if (a[1] == b[0] && a[0] == b[1])return false;
        if (a[0] == b[2] && a[2] == b[0])return false;
        else return true;
    }
    else if (n == 2)
    {
        if (a[1] == b[0] && a[0] == b[1])return false;
        else return true;
    }
    else if (n == 1) return true;
}

bool isEqual(int a[3], int b[3])
{
    if (a[0] == b[0] && a[2] == b[2] && a[1] == b[1])return true;
    else return false;
}

int bfs()
{
    conn[0][0] = 1;
    visited[start[0]][start[1]][start[2]] = true;
    queIns(start,0);
    int curStatus[3],curStep;
    int nextStatus[3];
    while (!queIsEmpty())
    {
        curStep = getQueFront()[0];
        curStatus[0] = getQueFront()[1];
        curStatus[1] = getQueFront()[2];
        curStatus[2] = getQueFront()[3];
        //cout << curStatus[0] << " " << curStatus[1] << " " << curStatus[2] << " " << endl;
        quePop();
        if (isEqual(curStatus, goal)) 
            return curStep;
        //TODO:处理n！=3时此处的取值。
        for (int i = 1; i <= conn[curStatus[0]][0]; ++i)
        {
            nextStatus[0] = conn[curStatus[0]][i];
            for (int j = 1; j <= conn[curStatus[1]][0]; ++j)
            {
                nextStatus[1] = conn[curStatus[1]][j];
                for (int k = 1; k <= conn[curStatus[2]][0]; ++k)
                {
                    nextStatus[2] = conn[curStatus[2]][k];
                    if (isOverlap(nextStatus))continue;//若点重叠则跳过
                    if (!isAvalible(nextStatus, curStatus)) continue;//若互穿则跳过
                    if (visited[nextStatus[0]][nextStatus[1]][nextStatus[2]]) continue;//若已经经过这个状态则跳过
                    //若状态可行
                    visited[nextStatus[0]][nextStatus[1]][nextStatus[2]]=true;
                    queIns(nextStatus, curStep + 1);
                }
            }
        }
    }
    return -1;
}

int dh[] = {1,-1,0,0,0};
int dw[] = {0,0,1,-1,0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> w >> h >> n&&w != 0)
    {
        init();
        int tnum = 1;
        for (int i = 1; i <= h; ++i)
        {
            string line;
            while (line == "")getline(cin, line);
            for (int j = 1; j <= w; ++j)
            {
                G[i][j]=line[j-1];
                if (G[i][j] != '#') num[i][j] = tnum++;
                if (G[i][j] == 'A')goal[0] = num[i][j];
                else if (G[i][j] == 'B')goal[1] = num[i][j];
                else if (G[i][j] == 'C')goal[2] = num[i][j];
                else if (G[i][j] == 'a')start[0] = num[i][j];
                else if (G[i][j] == 'b')start[1] = num[i][j];
                else if (G[i][j] == 'c')start[2] = num[i][j];
            }
        }
        for (int i = 1; i <= h; ++i)
        {
            for (int j = 1; j <= w; ++j)
            {
                if (G[i][j] == '#')continue;
                int tpos = 1, ti, tj;
                for (int k = 0; k < 5; ++k)
                {
                    ti = i + dh[k];
                    tj = j + dw[k];
                    if (ti > 0 && ti <= h&&tj > 0 && tj <= w&&G[ti][tj] != '#')
                    {
                        conn[num[i][j]][0] += 1;
                        conn[num[i][j]][tpos++] = num[ti][tj];
                    }
                }
            }
        }
        int step = bfs();
        cout << step << endl;
    }
}

　　回头有时间还是写一下双向BFS和A*试试吧。