poj 1128 构图+dfs+拓扑排序

题意：

给出一个堆叠后的表示，求从底向上的图片的堆叠顺序。

具体描述看题目。

解析：

先构图，然后dfs+拓扑

代码：

#pragma comment(linker, "/STACK:1677721600")#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstdlib>#include <cstring>#include <climits>#include <cassert>#include <iostream>#include <algorithm>#define pb push_back#define mp make_pair#define LL long long#define lson lo,mi,rt<<1#define rson mi+1,hi,rt<<1|1#define Min(a,b) ((a)<(b)?(a):(b))#define Max(a,b) ((a)>(b)?(a):(b))#define mem(a,b) memset(a,b,sizeof(a))#define FIN freopen("in.txt", "r", stdin)#define FOUT freopen("out.txt", "w", stdout)#define rep(i,a,b) for(int i=(a); i<=(b); i++)#define dec(i,a,b) for(int i=(a); i>=(b); i--)using namespace std;const int mod = 1e9 + 7;const double eps = 1e-8;const double ee = exp(1.0);const int inf = 0x3f3f3f3f;const int maxn = 30 + 10;const double pi = acos(-1.0);int readT(){    char c;    int ret = 0,flg = 0;    while(c = getchar(), (c < '0' || c > '9') && c != '-');    if(c == '-') flg = 1; else ret = c ^ 48;    while( c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);    return flg ? - ret : ret;}LL readTL(){    char c;    int flg = 0;    LL ret = 0;    while(c = getchar(), (c < '0' || c > '9') && c != '-');    if(c == '-') flg = 1; else ret = c ^ 48;    while( c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);    return flg ? - ret : ret;}char maze[maxn][maxn];int r, c;int word_num;int inDegree[maxn];int g[maxn][maxn];bool vis[maxn];int lx, rx, uy, dy;void search_bound(int x, int y){    char ch = maze[x][y];    lx = uy = inf;    rx = dy = -inf;    rep(i, 0, r - 1)    {        rep(j, 0, c - 1)        {            if (maze[i][j] == ch)            {                uy = Min(uy, i);                dy = Max(dy, i);                lx = Min(lx, j);                rx = Max(rx, j);            }        }    }}void build_map(){    word_num = 0;    rep(i, 0, r - 1)    {        rep(j, 0, c - 1)        {            char ch = maze[i][j];            if (ch == '.' || inDegree[ch - 'A'] != -1)                continue;            word_num++;            inDegree[ch - 'A'] = 0;            search_bound(i, j);            rep(k, lx, rx)            {                if (maze[uy][k] != ch)                    g[ch - 'A'][maze[uy][k] - 'A'] = 1;                if (maze[dy][k] != ch)                    g[ch - 'A'][maze[dy][k] - 'A'] = 1;            }            rep(k, uy, dy)            {                if (maze[k][lx] != ch)                    g[ch - 'A'][maze[k][lx] - 'A'] = 1;                if (maze[k][rx] != ch)                    g[ch - 'A'][maze[k][rx] - 'A'] = 1;            }        }    }    rep(i, 0, 25)    {        rep(j, 0, 25)        {            if (g[i][j])            {                inDegree[j]++;            }        }    }}void dfs(int dep, char ans[]){    if (dep == word_num)    {        ans[dep] = '\0';        printf("%s\n", ans);        return ;    }    rep(i, 0, 25)    {        if (inDegree[i] == 0 && !vis[i])        {            ans[dep] = i + 'A';            vis[i] = true;            rep(j, 0, 25)            {                if (g[i][j])                    --inDegree[j];            }            dfs(dep + 1, ans);            vis[i] = false;            rep(j, 0, 25)            {                if (g[i][j])                    ++inDegree[j];            }        }    }}int main(){    #ifdef LOCAL    FIN;    #endif // LOCAL    while (~scanf("%d%d", &r, &c))    {        rep(i, 0, r - 1)        {            scanf("%s", maze[i]);        }        mem(inDegree, -1);        mem(g, 0);        build_map();        mem(vis, false);        char ans[maxn];        dfs(0, ans);    }    return 0;}