Codeforces Round #292 (Div. 1) B. Drazil and Tiles (类似拓扑)

题目链接：http://codeforces.com/problemset/problem/516/B

一个n*m的方格，'*'不能填。给你很多个1*2的尖括号，问你是否能用唯一填法填满方格。

类似topsort，'.'与上下左右的'.'，的相连。从度为1的点作为突破口。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 2e3 + ;
 int n, m, tx[] = {-, , , }, ty[] = {, -, , };
 char str[N][N];
 int du[N*N];
 vector <int> G[N*N];
 inline bool judge(int x, int y) {
     if(x >=  && x < n && y >=  && y < m && str[x][y] == '.')
         return true;
     return false;
 }
 inline void get(int x, int y) {
     for(int i =  ; i <  ; ++i) {
         int xx = x + tx[i], yy = y + ty[i];
         if(judge(xx, yy)) {
             G[x*m + y].push_back(xx*m + yy);
             du[x*m + y]++;
         }
     }
 }
 inline void change(int x, int y) {
     if(m == ) {
         if(x < y) {
             str[x][] = '^';
             str[y][] = 'v';
         } else {
             str[x][] = 'v';
             str[y][] = '^';
         }
     } else {
         if(x - y == -m) {
             str[x/m][x%m] = '^';
             str[y/m][y%m] = 'v';
         } else if(x - y == m) {
             str[x/m][x%m] = 'v';
             str[y/m][y%m] = '^';
         } else if(x - y == -) {
             str[x/m][x%m] = '<';
             str[y/m][y%m] = '>';
         } else {
             str[x/m][x%m] = '>';
             str[y/m][y%m] = '<';
         }
     }
 }

 int main()
 {
     int cnt = , op = ;
     scanf("%d %d", &n, &m);
     for(int i = ; i < n; ++i) {
         scanf("%s", str[i]);
     }
     for(int i = ; i < n; ++i) {
         for(int j = ; j < m; ++j) {
             if(judge(i, j)) {
                 get(i, j);
                 ++op;
             }
         }
     }
     queue <int> que;
     for(int i = ; i < n*m; ++i) {
         if(du[i] == )
             que.push(i);
     }
     while(!que.empty()) {
         int u = que.front();
         que.pop();
         du[u]--;
         for(int i = ; i < G[u].size(); ++i) {
             int v = G[u][i];
             du[v]--;
             if(du[v] > ) {
                 cnt++;
                 for(int j = ; j < G[v].size(); ++j) {
                     du[G[v][j]]--;
                     if(du[G[v][j]] == ) {
                         que.push(G[v][j]);
                     }
                 }
                 du[v] = ;
                 change(u, v);
             } else if(du[v] == ) {
                 cnt++;
                 change(u, v);
             }
         }
     }
     if(cnt *  == op) {
         for(int i = ; i < n; ++i) {
             printf("%s\n", str[i]);
         }
     } else {
         printf("Not unique\n");
     }
     return ;
 }