The Chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 466 Accepted Submission(s): 130
Problem Description
Mr Li likes playing chess very much. One day he made a simplified version of the game. The chess board is relatively smaller. He will first place only three kinds of chesses on his own behalf, namely the Chariot("ju"), the Horse("ma")
and the Cannon("pao"). Then he placed some chesses on the opposite‘s behalf. There will be one and only one chess of the opposite marked as the General("shuai"). Mr Li wants to kill the General as soon as possible without killing other chesses.
All chesses will move in the same way as in real chess games(The chinese chess).
To make the game easier, the opposite‘s chesses will stand still.
Input
For each test case,the first line contains two integers n and m(2<=n,m<=10) indicating the size of the borad. The next n lines, containing m characters each, describe the board. ‘C‘, ‘M‘ and ‘P‘ stand for the Chariot, the Horse and
the Cannon on Mr Li‘s behalf. ‘S‘ stands for the General of the opposite while ‘D‘ stands for other chesses. Cells marked as ‘.‘ are empty.
Process to the end of file.
Output
For each test case, first print a line saying "Scenario #k", where k is the number of the test case.Then,if Mr Li can kill the General, print the least number of steps needed. Otherwise print the sentence "OH!That‘s impossible!".Print
a blank line after each test case, even after the last one.
Sample Input
5 5 ..DSD ...D. C.... P.D.. ...M. 7 7 .DDSDD. ..DDD.. ...D... .....P. .C..... ...M... .......
Sample Output
Scenario #1 2 Scenario #2 OH!That‘s impossible!
分析：这是一个很好的搜索问题，不知为什么用cin就wa，换成%s就ac了，坑了一天，坑急眼了连饭都没吃。
广搜就是遍历车，马，炮所有的位置状态，开一个6维数组标记3个元素的位置，车，马比较好遍历，车4个方向，马8个·方向，炮除了可以和车一样走外
还可以，还可以“隔山打牛”。注意：马可以被拨马腿，炮不能走到帅的位置上。
代码示例：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
typedef struct
{
int xj,xm,xp;
int yj,ym,yp;
int step;
}node;
int map[11][11],dis[11][11][11][11][11][11];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int dit[8][2]={{2,1},{1,2},{-2,1},{1,-2},{2,-1},{-1,2},{-2,-1},{-1,-2}};
int dic[8][2]={{1,0},{0,1},{-1,0},{0,-1},{1,0},{0,1},{-1,0},{0,-1}};
int N,M,ax,ay,bx,by,cx,cy,X,Y;
int min(int a,int b)
{
return a<b?a:b;
}
int max(int a,int b)
{
return a>b?a:b;
}
int bfs()
{
int num,h,t;
node fir,nex;
fir.xj=ax,fir.xm=bx,fir.xp=cx;
fir.yj=ay,fir.ym=by,fir.yp=cy;
fir.step=0;
dis[ax][ay][bx][by][cx][cy]=1;
queue<node>Q;
Q.push(fir);
while(!Q.empty())
{
fir=Q.front();
Q.pop();
map[fir.xj][fir.yj]=1;
map[fir.xm][fir.ym]=1;
map[fir.xp][fir.yp]=1;
for(int i=0;i<8;i++)
{
nex=fir;
nex.xm=fir.xm+dit[i][0];
nex.ym=fir.ym+dit[i][1];
nex.step=fir.step+1;
if((fir.xm+dic[i][0]==X)&&(fir.ym+dic[i][1]==Y))continue;
if(map[fir.xm+dic[i][0]][fir.ym+dic[i][1]])continue;
if(nex.xm>=1&&nex.xm<=N&&nex.ym>=1&&nex.ym<=M&&!map[nex.xm][nex.ym]&&!dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp])
{
if(nex.xm==X&&nex.ym==Y)
{
return nex.step;
}
dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp]=1;
Q.push(nex);
}
}
for(int i=0;i<4;i++)
{
nex=fir;
nex.xj=fir.xj+dir[i][0];
nex.yj=fir.yj+dir[i][1];
nex.step=fir.step+1;
while(1)
{
if(nex.xj>=1&&nex.xj<=N&&nex.yj>=1&&nex.yj<=M&&!map[nex.xj][nex.yj]&&!dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp])
{
if(nex.xj==X&&nex.yj==Y)
{
return nex.step;
}
dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp]=1;
Q.push(nex);
}
else
{
break;
}
nex.xj+=dir[i][0];
nex.yj+=dir[i][1];
}
}
for(int i=0;i<4;i++)
{
nex=fir;
nex.xp=fir.xp+dir[i][0];
nex.yp=fir.yp+dir[i][1];
nex.step=fir.step+1;
num=0;
while(1)
{
if(!(nex.xp>=1&&nex.xp<=N&&nex.yp>=1&&nex.yp<=M))break;
if(map[nex.xp][nex.yp])num++;
if(num>1)break;
if(nex.xp==X&&nex.yp==Y&&num==0)
break;
if(nex.xp==X&&nex.yp==Y&&num==1)
{
// printf("<%d,%d>\n",nex.xp,nex.yp);
return nex.step;
}
if(!map[nex.xp][nex.yp]&&!dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp]&&num==0)
{
dis[nex.xj][nex.yj][nex.xm][nex.ym][nex.xp][nex.yp]=1;
Q.push(nex);
}
nex.xp+=dir[i][0];
nex.yp+=dir[i][1];
}
}
map[fir.xj][fir.yj]=0;
map[fir.xm][fir.ym]=0;
map[fir.xp][fir.yp]=0;
}
return -1;
}
int main()
{
char c[21][21];
int l=0;
while(~scanf("%d%d",&N,&M))
{
memset(dis,0,sizeof(dis));
for(int i=1;i<=N;i++)
scanf("%s",c[i]+1);
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
{
//cin>>c;
if(c[i][j]==‘D‘)
{
map[i][j]=1;
}
else
{
map[i][j]=0;
}
if(c[i][j]==‘S‘)
{
X=i,Y=j;
continue;
}
if(c[i][j]==‘C‘)
{
ax=i,ay=j;
continue;
}
if(c[i][j]==‘M‘)
{
bx=i,by=j;
continue;
}
if(c[i][j]==‘P‘)
{
cx=i,cy=j;
}
}
int time=bfs();
l++;
printf("Scenario #%d\n",l);
if(time==-1)
{
printf("OH!That‘s impossible!\n\n");
}
else
printf("%d\n\n",time);
}
return 0;
}