我试图在spoj上解决广泛的第一个搜索问题。问题陈述如下： n×n方形单元阵列的等分是阵列中n ^ 2个单元的恰好n组的分区，每个单元具有n个连续单元。当两个单元具有共同侧时，它们是连续的。 问题链接是：http://www.spoj.com/problems/EQDIV/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<sstream>
#include<queue>
using namespace std;
#define pii pair< int, int>
int n, m, grid[105][105];
char inp[101 * 101];
bool inRange(int i, int j)
{
m = n;
return (i >= 0 && i < n && j >= 0 && j < m);
}
int main()
{
int i, j, comp;
scanf("%d", &n);
while (n != 0)
{
if (n == 1)
{
printf("good\n");
scanf("%d", &n);
continue;
}
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
grid[i][j] = 0;
}
}
int a, b, start1[n], start2[n], print = 0, c, d;
queue< pii > Q;
pii p;
getchar();
for (i = 1; i <= n - 1; i++)
{
gets(inp);
stringstream ss(inp);
for (j = 1; j <= n; j++)
{
ss >> a >> b;
if (j == 1)
{
start1[i - 1] = a;
start2[i - 1] = b;
//printf("%d %d\n",start1[i-1],start2[i-1]);
}
grid[a - 1][b - 1] = i;
}
}
for (int x = 0; x < n; x++)
{
int count = 1;
if (x == n - 1)
{
comp = 0;
p.first = c;
p.second = d;
Q.push(p);
}
else
{
comp = x + 1;
p.first = start1[x] - 1;
p.second = start2[x] - 1;
Q.push(p);
}
while (!Q.empty())
{
p = Q.front();
i = p.first;
j = p.second;
Q.pop();
grid[i][j] = -1;
if (x != n - 1)
{
if (inRange(i + 1, j) && grid[i + 1][j] == 0)
{
c = i + 1;
d = j;
}
if (inRange(i - 1, j) && grid[i - 1][j] == 0)
{
c = i - 1;
d = j;
}
if (inRange(i, j + 1) && grid[i][j + 1] == 0)
{
c = i;
d = j + 1;
}
if (inRange(i, j - 1) && grid[i][j - 1] == x + 1)
{
c = i;
d = j - 1;
}
}
if (inRange(i + 1, j) && grid[i + 1][j] == comp)
{
p.first = i + 1;
p.second = j;
Q.push(p);
count += 1;
}
if (inRange(i - 1, j) && grid[i - 1][j] == comp)
{
p.first = i - 1;
p.second = j;
Q.push(p);
count += 1;
}
if (inRange(i, j + 1) && grid[i][j + 1] == comp)
{
p.first = i;
p.second = j + 1;
Q.push(p);
count += 1;
}
if (inRange(i, j - 1) && grid[i][j - 1] == comp)
{
p.first = i;
p.second = j - 1;
Q.push(p);
count += 1;
}
}
if (count != n)
{
print = 1;
printf("wrong\n");
break;
}
}
if (print == 0)
printf("good\n");
//printf("%d %d\n",c,d);
scanf("%d", &n);
}
}