超级麻烦的一题,开始还理解错题目意思了。。。!看着udebug的数据才明白,重读一遍题目。。。
三个机器人移动一致,碰到障碍物的机器人就无法向障碍物方向移动,机器人自己也算障碍物。麻烦在机器人当作障碍物的地方。
标记状态BFS暴力
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <math.h>
#include <ctype.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 100 + 5;
int t,n,kase = 0;
int dirx[] = {1,0,0,-1};
int diry[] = {0,-1,1,0};
char mp[15][15];
int stx[3],sty[3];
int edx[3],edy[3],edp;
int d[12][12][12][12][12][12];
struct Node
{
int ax,ay;
int bx,by;
int cx,cy;
Node(){}
Node(int a, int b, int c, int d, int e, int f)
{
ax = a; ay = b;
bx = c; by = d;
cx = e; cy = f;
}
};
int bfs(int ax, int ay, int bx, int by, int cx, int cy)
{
queue<Node> Q;
memset(d, -1, sizeof(d));
Node s(ax,ay,bx,by,cx,cy);
Q.push(s);
d[ax][ay][bx][by][cx][cy] = 0;
while(!Q.empty())
{
Node tmp = Q.front(); Q.pop();
int ax = tmp.ax, ay = tmp.ay;
int bx = tmp.bx, by = tmp.by;
int cx = tmp.cx, cy = tmp.cy;
for(int i = 0; i < 4; i++)
{
int nax = ax + dirx[i], nay = ay + diry[i];
int fa = 0,fb = 0,fc = 0;
if(nax < 0 || nay < 0 || nax >= n || nay >= n || mp[nax][nay] == '#')
{
nax = ax, nay = ay; fa = 1;
}
int nbx = bx + dirx[i], nby = by + diry[i];
if(nbx < 0 || nby < 0 || nbx >= n || nby >= n || mp[nbx][nby] == '#' )
{
nbx = bx, nby = by; fb = 1;
}
int ncx = cx + dirx[i], ncy = cy + diry[i];
if(ncx < 0 || ncy < 0 || ncx >= n || ncy >= n || mp[ncx][ncy] == '#' )
{
ncx = cx, ncy = cy; fc = 1;
}
if(fa && nbx == nax && nby == nay) nbx = bx, nby = by, fb = 1;
if(fa && ncx == nax && ncy == nay) ncx = cx, ncy = cy, fc = 1;
if(fb && nax == nbx && nay == nby) nax = ax, nay = ay, fa = 1;
if(fb && ncx == nbx && ncy == nby) ncx = cx, ncy = cy, fc = 1;
if(fc && nbx == ncx && nby == ncy) nbx = bx, nby = by, fb = 1;
if(fc && nax == ncx && nay == ncy) nax = ax, nay = ay, fa = 1;
if(fa && nbx == nax && nby == nay) nbx = bx, nby = by, fb = 1;
if(fa && ncx == nax && ncy == nay) ncx = cx, ncy = cy, fc = 1;
if(fb && nax == nbx && nay == nby) nax = ax, nay = ay, fa = 1;
if(fb && ncx == nbx && ncy == nby) ncx = cx, ncy = cy, fc = 1;
if(fc && nbx == ncx && nby == ncy) nbx = bx, nby = by, fb = 1;
if(fc && nax == ncx && nay == ncy) nax = ax, nay = ay, fa = 1;
if(d[nax][nay][nbx][nby][ncx][ncy] >= 0) continue;
int ans = d[ax][ay][bx][by][cx][cy] + 1;
d[nax][nay][nbx][nby][ncx][ncy] = ans;
if(mp[nax][nay] == 'X' && mp[nbx][nby] == 'X' && mp[ncx][ncy] == 'X')
return ans;
Q.push(Node(nax,nay,nbx,nby,ncx,ncy));
}
}
return -1;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%s", mp[i]);
edp=0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
if(mp[i][j] == 'A')
{
mp[i][j] = '.';
stx[0] = i; sty[0] = j;
}
else if(mp[i][j] == 'B')
{
mp[i][j] = '.';
stx[1] = i; sty[1] = j;
}
else if(mp[i][j] == 'C')
{
mp[i][j] = '.';
stx[2] = i; sty[2] = j;
}
}
}
int ans = bfs(stx[0],sty[0],stx[1],sty[1],stx[2],sty[2]);
printf("Case %d: ", ++kase);
if(ans == -1)
printf("trapped
");
else
printf("%d
", ans);
}
return 0;
}