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USACO 4.4 Frame Up(拓扑排序)

2013-05-17 10:39 204 查看
终于结束了第4章。。倒数第二个题,卡了2个月。。。



这个题,实现起来,还挺麻烦的,看懂题意,建好图,就是裸的拓扑排序了。

/*
ID:cuizhe
LANG: C++
TASK: frameup
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <iostream>
using namespace std;
int lx[30],ly[30],rx[30],ry[30],flag[31],pre[31];
char str[101][101];
int p[30][30],o[30],key[30];
void dfs(int x,int step)
{
int i,z,j;

pre[step] = x;
z = 0;
for(i = 0; i < 26; i ++)
{
if(!flag[i]&&o[i] == 0&&key[i])
{
for(j = 0; j < 26; j ++)
{
if(p[i][j])
o[j] --;
}
flag[i] = 1;
dfs(i,step+1);
flag[i] = 0;
for(j = 0; j < 26; j ++)
{
if(p[i][j])
o[j] ++;
}
z = 1;
}
}
if(!z)
{
for(i = 1; i <= step; i ++)
printf("%c",pre[i]+'A');
printf("\n");
return ;
}
}
int main()
{
int n,m,i,j;
freopen("frameup.in","r",stdin);
freopen("frameup.out","w",stdout);
scanf("%d%d",&n,&m);
for(i = 0; i < 26; i ++)
{
lx[i] = 100000;
ly[i] = 100000;
rx[i] = -1;
ry[i] = -1;
}
for(i = 0; i < n; i ++)
{
scanf("%s",str[i]);
}
for(i = 0; i < n; i ++)
{
for(j = 0; j < m; j ++)
{

if(str[i][j] != '.')
{
key[str[i][j]-'A'] = 1;
lx[str[i][j]-'A'] = min(lx[str[i][j]-'A'],i);
ly[str[i][j]-'A'] = min(ly[str[i][j]-'A'],j);
rx[str[i][j]-'A'] = max(rx[str[i][j]-'A'],i);
ry[str[i][j]-'A'] = max(ry[str[i][j]-'A'],j);
}
}
}
for(i = 0; i < 26; i ++)
{
if(key[i])
{
for(j = lx[i]; j <= rx[i]; j ++)
{
if(str[j][ly[i]] != 'A' + i&&!p[i][str[j][ly[i]]-'A'])
{
p[i][str[j][ly[i]]-'A'] = 1;
o[str[j][ly[i]]-'A'] ++;
}
}
for(j = lx[i]; j <= rx[i]; j ++)
{
if(str[j][ry[i]] != 'A' + i&&!p[i][str[j][ry[i]]-'A'])
{
p[i][str[j][ry[i]]-'A'] = 1;
o[str[j][ry[i]]-'A'] ++;
}
}
for(j = ly[i]; j <= ry[i]; j ++)
{
if(str[lx[i]][j] != 'A' + i&&!p[i][str[lx[i]][j]-'A'])
{
p[i][str[lx[i]][j]-'A'] = 1;
o[str[lx[i]][j]-'A'] ++;
}
}
for(j = ly[i]; j <= ry[i]; j ++)
{
if(str[rx[i]][j] != 'A' + i&&!p[i][str[rx[i]][j]-'A'])
{
p[i][str[rx[i]][j]-'A'] = 1;
o[str[rx[i]][j]-'A'] ++;
}
}
}
}
for(i = 0; i < 26; i ++)
{
if(o[i] == 0&&key[i])
{
flag[i] = 1;
for(j = 0; j < 26; j ++)
{
if(p[i][j])
o[j] --;
}
dfs(i,1);
for(j = 0; j < 26; j ++)
{
if(p[i][j])
o[j] ++;
}
flag[i] = 0;
}
}
return 0;
}
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