UVA 1572 SelfAssembly（图论模型+拓扑排序）

题意：判断利用给出的正方形是否能拼接出无限延伸的结构。

分析：正方形上的字母看做点，正方形看做有向边。

例如：

若上下两个正方形能拼接，需要B+~C+是个有向边。

对输入的处理是：把A+,A-分别映射成2n+1,2n，利用（2n）^1 = 2n+1 , （2n+1）^1 = 2n 的性质处理有向边。

若存在有向环则unbounded，即不存在拓扑排序

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 52 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[10];
int pic[MAXN][MAXN];
int vis[MAXN];
priority_queue<int> q;
int get_id(char a, char b){
    return b == '+' ? 2 * (a - 'A') + 1 : 2 * (a - 'A');
}
bool dfs(int x){
    vis[x] = -1;
    for(int i = 0; i < 52; ++i){
        if(pic[x][i]){
            if(vis[i] == -1) return false;
            else if(!vis[i] && !dfs(i)) return false;
        }
    }
    vis[x] = 1;
    return true;
}
bool toposort(){
    for(int i = 0; i < 52; ++i){
        if(!vis[i] && !dfs(i)){
            return false;
        }
    }
    return true;
}
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        memset(pic, 0, sizeof pic);
        memset(vis, 0, sizeof vis);
        while(n--){
            scanf("%s", s);
            for(int i = 0; i < 4; ++i){//处理除了00之外的点之间的有向边的关系
                for(int j = 0; j < 4; ++j){
                    if(i != j && s[i * 2] != '0' && s[j * 2] != '0'){
                        int from = get_id(s[i * 2], s[i * 2 + 1]) ^ 1;
                        int to = get_id(s[j * 2], s[j * 2 + 1]);
                        pic[from][to] = 1;
                    }
                }
            }
        }
        bool ok = toposort();
        if(ok){
            printf("bounded\n");
        }
        else{
            printf("unbounded\n");
        }
    }
    return 0;
}