POJ 1324(BFS + 状态压缩)

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题意：给你一条蛇，要求一以最少的步数走到1,1

思路：

最开始一直没想到应该怎样保存状态，后来发现别人用二进制保存蛇的状态，即每两个节点之间的方向和头节点，二进制最多14位（感觉状态保存都能扯到二进制）。然后就是bfs

问题：

1.最开始完全没想到状态压缩的问题

2.感觉现在做题太急，做题没有足够的思考，思路不清晰便开始写，导致在过程中经常崩盘。

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <cstdlib>#include <queue>#include <algorithm>typedef long long ll;using namespace std;int flag;const int maxn = 25;const int ma = (1<<14)+10;int vis[maxn][maxn];int sta[maxn][maxn][ma];struct node{//    int snake[15][2];    int x,y;    int step;    int state;};node cur;int t,n,m;int dir[4][2] = {{-1,0},{0,-1},{1,0},{0,1}};int matri[20][2];int get_state()   //蛇的状态{    int dir;    int stat = 0;    for(int i = t; i > 1; i --)    {        int x = matri[i][0]-matri[i-1][0];        int y = matri[i][1]-matri[i-1][1];        if(x == -1 && y == 0)            dir = 0;        if(x == 0 && y == -1)            dir = 1;        if(x == 1 && y == 0)            dir =2 ;        if(x == 0 && y == 1)            dir = 3;        stat <<= 2;        stat = stat |dir;    }    return stat;}bool judge(int x,int y,int x1,int y1,int  state)        //是否吃自己{    int s;    for(int i = 1; i < t; i++)    {        s = 3;        s = state & s;            //取最后两位        state = state >> 2;        if(x == x1+dir[s][0] && y == y1+dir[s][1])            return true;        x1  = x1 + dir[s][0];        y1  = y1 + dir[s][1];    }    return false ;}int get_next(int i,int state)       //去掉高位，输入低位{    int x = 0-dir[i][0];    int y = 0-dir[i][1];    int dir;    int k = (1 << ((t - 1) << 1)) - 1;    if(x == -1 && y == 0)        dir = 0;    if(x == 0 && y == -1)        dir = 1;    if(x == 1 && y == 0)        dir =2 ;    if(x == 0 && y == 1)        dir = 3;    state <<= 2;    state |= dir;    state = state & k;        //去掉最高位    return state;}void bfs(){    queue<node>q;    cur.step = 0;    sta[cur.x][cur.y][cur.state] = 1;    q.push(cur);    node fo;    while(!q.empty())    {        fo = q.front();        q.pop();        for(int i = 0; i < 4; i++)        {            int x = fo.x+dir[i][0];            int y = fo.y+dir[i][1];            int ts = get_next(i,fo.state);            if(x < 1 || x > n || y <1 || y > m || vis[x][y] || judge(x,y,fo.x,fo.y,fo.state) || sta[x][y][ts])                continue;            if(x == 1 && y == 1)            {                printf("%d\n",fo.step+1);                flag = 1;                return;            }            node tmp;            tmp.x =x ;            tmp.y = y;            tmp.step = fo.step + 1;//            for(int i = t; i >= 1; i--)//            {//                tmp.snake[i][0] = tt.snake[i-1][0];//                tmp.snake[i][1] = tt.snake[i-1][1];//            }            tmp.state = ts;            sta[x][y][tmp.state] = 1;            q.push(tmp);        }    }}int main(){    int cas= 1;    while(scanf("%d%d%d",&n,&m,&t) != EOF)    {        memset(vis,0,sizeof(vis));        memset(sta,0,sizeof(sta));        if(n == 0 && m == 0 && t == 0)            break;        for(int i = 1; i <= t; i++)            scanf("%d%d",&matri[i][0],&matri[i][1]);        int a,b,k;        scanf("%d",&k);        for(int i = 0; i < k; i++)        {            scanf("%d%d",&a,&b);            vis[a][b] = 1;        }        cur.x = matri[1][0];        cur.y = matri[1][1];        cur.state = get_state();        flag = 0;        printf("Case %d: ",cas++);        if(cur.x == 1 && cur.y == 1)        {            printf("0\n");            continue;        }        bfs();        if(!flag)            printf("-1\n");    }}

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