USACO Frame Up 矩阵覆盖 拓扑排序

　　矩阵覆盖的题， 可以转化为拓扑排序。写这个拓扑排序废了好大的劲以后好好看看。代码如下：

/*
    ID: m1500293
    LANG: C++
    PROG: frameup
*/

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

using namespace std;
int h, w;
int x1[30], y1[30], x2[30], y2[30];
int ex[30];     //有没有出现
char Map[55][55];
int indeg[30];
int d[30][30];

int ans[30], nans;

void dfs()       //很精髓的拓扑排序
{
    bool flog = 0;
    for(int i=1; i<=26; i++) if(ex[i])
    {
        flog = 1;
        break;
    }
    if(!flog)
    {
        for(int i=0; i<nans; i++) printf("%c", ans[i]+'@');
        printf("\n");
        return ;
    }
    for(int u=1; u<=26; u++) if(ex[u] && !indeg[u])
    {
        ex[u] = 0;
        ans[nans++] = u;
        for(int v=1; v<=26; v++) if(d[u][v]) indeg[v]--;
        dfs();
        ex[u] = 1;
        nans--;
        for(int v=1; v<=26; v++) if(d[u][v]) indeg[v]++;
    }
}

int main()
{
    freopen("frameup.in", "r", stdin);
    freopen("frameup.out", "w", stdout);
    scanf("%d%d", &h, &w);
    for(int i=1; i<=h; i++)
    {
        scanf("%s", Map[i]+1);
        for(int j=1; j<=w; j++)
        {
            if(Map[i][j] == '.') continue;
            int id = Map[i][j] - '@';
            if(!x1[id] || x1[id]>i) x1[id] = i;
            if(!y1[id] || y1[id]>j) y1[id] = j;
            if(!x2[id] || x2[id]<i) x2[id] = i;
            if(!y2[id] || y2[id]<j) y2[id] = j;
            ex[id] = 1;
        }
    }
    for(int u=1; u<=26; u++) if(ex[u])
    {
        //横着扫 x1[u] x2[u] y1[u] -> y2[u]
        for(int y=y1[u]; y<=y2[u]; y++)
        {
            int id = Map[x1[u]][y] - '@';
            if(id != u && !d[u][id]) d[u][id] = 1, indeg[id]++;
            id = Map[x2[u]][y] - '@';
            if(id != u && !d[u][id]) d[u][id] = 1, indeg[id]++;
        }
        for(int x=x1[u]; x<=x2[u]; x++)
        {
            int id = Map[x][y1[u]] - '@';
            if(id != u && !d[u][id]) d[u][id] = 1, indeg[id]++;
            id = Map[x][y2[u]] - '@';
            if(id != u && !d[u][id]) d[u][id] = 1, indeg[id]++;
        }
    }
    nans = 0;
    dfs();
    return 0;
}