模板题
按顺序BFS一边就可以判断结果了
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <string>
#include <math.h>
#include <bitset>
#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 = 15;
char mp[N][N];
int d[N][N],n;
int dirx[] = {0,1,0,-1};
int diry[] = {1,0,-1,0};
struct Node
{
int id;
int x,y;
Node(){}
Node(int a, int b, int c){
id = a, x = b, y = c;
}
};
bool cmp(Node a, Node b)
{
return a.id<b.id;
}
int t,kase=0;
int bfs(int i, int j, int ei, int ej)
{
queue<P> Q;
memset(d, -1, sizeof(d));
Q.push(P(i,j));
d[i][j] = 0;
while(!Q.empty())
{
P tmp = Q.front(); Q.pop();
int x = tmp.first, y = tmp.second;
for(int i = 0; i < 4; i++)
{
int nx = x + dirx[i], ny = y + diry[i];
if(nx < 0 || ny < 0 || nx >= n || ny >= n) continue;
if(mp[nx][ny] != '.') continue;
if(d[nx][ny] >= 0) continue;
if(nx == ei && ny == ej) return d[x][y] + 1;
d[nx][ny] = d[x][y] + 1;
Q.push(make_pair(nx,ny));
}
}
return -1;
}
int main()
{
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%s", mp[i]);
}
vector<Node> p;
p.clear();
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
if(isupper(mp[i][j]))
p.push_back(Node(mp[i][j]-'A',i,j));
}
sort(p.begin(),p.end(),cmp);
int ans = 0, flag = 0;
for(int i = 1; i < p.size(); i++)
{
mp[p[i].x][p[i].y] = '.';
mp[p[i-1].x][p[i-1].y] = '.';
int tm = bfs(p[i-1].x, p[i-1].y, p[i].x, p[i].y);
if(tm == -1)
flag = 1;
ans += tm;
}
printf("Case %d: ", ++kase);
if(flag)
printf("Impossible
");
else
printf("%d
", ans);
}
return 0;
}