POJ-1128 拓扑排序水题

由于题目不理解wa了几发。

大意是给一张图，是多个由单种字母组成的框叠成的。框的每条边都会至少给出一个字母（这个我开始没注意到，构图查找时想多了）。

输出叠放的顺序。

思路是先构出有向关系边图，然后拓扑排序输出。

拓扑排序是先找入度为0的点，然后删除该点（该点后续相邻点的入度减一）。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>

const int maxn = 31;
int map[31][31];
int re[31][31];
struct Inf
{
int l, r, t, d;
}Inf[28];
int m, n, cnt;
int du[28];
int vis[28];
int tot;
char str[32];
void dfs(int depth)
{
if(depth == tot){
for(int i = 0; i < tot; i++)printf("%c",str[i]);
printf("\n");
return;
}
for(int i = 0; i <= cnt; i++)
{
if(du[i] == 0)
{
str[depth] = i + 'A';
du[i] = -1;
for(int j = 0; j <= cnt; j++)
if(re[i][j]) du[j] --;
dfs(depth + 1);
du[i] = 0;
for(int j = 0; j <= cnt; j++)
if(re[i][j]) du[j] ++;
}
}
return;
}

int main()
{
while(~scanf("%d%d", &m, &n))
{
int i, j, tmp;
cnt = 0;
char str[31];
for(i = 0; i < 27; i++)
{
Inf[i].l = n;
Inf[i].r = 0;
Inf[i].t = m;
Inf[i].d = 0;
}
memset(re, 0, sizeof(re));
memset(vis, 0, sizeof(vis));
for(i = 0; i < m; i++)
{
scanf("%s", str);
for(j = 0; j < n; j++)
{
if(str[j] == '.')map[i][j] = -1;
else {
tmp = str[j] - 'A';
if(cnt < tmp)cnt = tmp;
map[i][j] = tmp;
vis[tmp] = 1;
if(tmp > cnt)cnt = map[i][j];
if(Inf[tmp].l > j)Inf[tmp].l = j;
if(Inf[tmp].r < j)Inf[tmp].r = j;
if(Inf[tmp].t > i)Inf[tmp].t = i;
if(Inf[tmp].d < i)Inf[tmp].d = i;
}
}
}
tot = 0;
for(i = 0; i <= cnt; i++) if(vis[i])tot++;
str[tot] = '\0';
for(i = 0; i <= cnt; i++)
{
if(vis[i])
for(j = Inf[i].l; j <= Inf[i].r; j++)
{
int top = Inf[i].t;
int down = Inf[i].d;
if(map[top][j] != -1 && map[top][j] != i)
re[i][map[top][j]] = 1;
if(map[down][j] != -1 && map[down][j] != i)
re[i][map[down][j]] = 1;
}
if(vis[i])
for(j = Inf[i].t; j <= Inf[i].d; j++)
{
int left = Inf[i].l;
int right = Inf[i].r;
if(map[j][left] != -1 && map[j][left] != i)
re[i][map[j][left]] = 1;
if(map[j][right] != -1 && map[j][right] != i)
re[i][map[j][right]] = 1;
}
}
for(i = 0; i <= cnt; i++)
if(vis[i])du[i] = 0;
else du[i] = -1;
for(i = 0; i <= cnt; i++)
{
for(j = 0; j <= cnt; j++)
if(re[i][j])du[j]++;
}
dfs(0);
}
return 0;
}