POJ 1178 Camelots(bfs)

题意：把这些所有的棋子移动到一个点，king可以自己走或者骑马走。

思路：预处理，枚举所有的点作为他们的集合点。。

先计算马走的步数，再计算king到最近马路的步数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>#include <stack>#include <map>using namespace std;const int  N = 8;int re[N][N][N][N];queue<int> que;bool inline oor(int x,int y){    if(x<0||x>=8) return false;    if(y<0||y>=8) return false;    return true;}int dx[] = {-2,-2,-1,-1,1,1,2,2};int dy[] = {1,-1,2,-2,2,-2,1,-1};void mindis(int x,int y){    while(!que.empty()) que.pop();    que.push((x<<5)|y);    re[x][y][x][y] = 0;    int e,ex,ey,tx,ty;    while(!que.empty())    {        e = que.front(); que.pop();        ex = e>>5;ey = e&15;        for(int i=0;i<8;i++)        {            tx = ex+dx[i],ty=ey+dy[i];            if((!oor(tx,ty))||(re[x][y][tx][ty]!=-1)) continue;            re[x][y][tx][ty] = re[x][y][ex][ey]+1;            que.push(tx<<5|ty);        }    }}void init(){    memset(re,-1,sizeof(re));    for(int i=0;i<N;i++)    for(int j=0;j<N;j++)    mindis(i,j);}char ch[256];bool visit[8][8];struct nod{    int x,y,dis;};queue <nod> qq;int dxx[]={0,0,1,1,1,-1,-1,-1};int dyy[]={-1,1,-1,0,1,-1,0,1};int kinv[8][8];int oper(int x,int y){    memset(visit,0,sizeof(visit));    visit[x][y] = 1;    while(!qq.empty()) qq.pop();    nod e,t;    int len = strlen(ch);    int ret = 0;    for(int i=2;i<len;i+=2)    {        e.x = ch[i]-'A';e.y= ch[i+1]- '1';        e.dis = re[x][y][e.x][e.y];        qq.push(e);        visit[e.x][e.y] = 1;        ret+=e.dis;    }    int tx,ty;    while(!qq.empty())    {        e = qq.front();qq.pop();        for(int i=0;i<8;i++)        {            tx=e.x+dx[i],ty=e.y+dy[i];            if((!oor(tx,ty))||visit[tx][ty]||(re[x][y][tx][ty]!=e.dis-1)) continue;            t.x = tx,t.y = ty;t.dis=e.dis-1;            visit[tx][ty] = 1;            qq.push(t);        }    }    while(!que.empty()) que.pop();    que.push((ch[0]-'A')<<5|(ch[1]-'1'));    memset(kinv,0,sizeof(kinv));    kinv[ch[0]-'A'][ch[1]-'1'] = 1;    while(!que.empty())    {        int ee = que.front();que.pop();        int ex = ee>>5,ey = ee&15;        if(visit[ex][ey]) {            return ret+ kinv[ex][ey] -1;        }        for(int i=0;i<8;i++)        {            int tx = ex+dxx[i],ty=ey+dyy[i];            if(!oor(tx,ty)||kinv[tx][ty]) continue;            kinv[tx][ty] = kinv[ex][ey]+1;            que.push(tx<<5|ty);        }    }    return 100;}void solve(){    int ans = 10000000;    for(int i=0;i<8;i++)    for(int j=0;j<8;j++)    ans = min(ans,oper(i,j));    printf("%d\n",ans);}int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    init();    scanf("%s",ch);    solve();    return 0;}