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。

  1 //#pragma comment(linker, "/STACK:102400000, 102400000")
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <cstdlib>
  5 #include <cstring>
  6 #include <cstdio>
  7 #include <vector>
  8 #include <queue>
  9 #include <cmath>
 10 #include <ctime>
 11 #include <list>
 12 #include <set>
 13 #include <map>
 14 using namespace std;
 15 typedef long long LL;
 16 typedef pair <int, int> P;
 17 const int N = 2e3 + 5;
 18 int n, m, tx[] = {-1, 0, 1, 0}, ty[] = {0, -1, 0, 1};
 19 char str[N][N];
 20 int du[N*N];
 21 vector <int> G[N*N];
 22 inline bool judge(int x, int y) {
 23     if(x >= 0 && x < n && y >= 0 && y < m && str[x][y] == ‘.‘)
 24         return true;
 25     return false;
 26 }
 27 inline void get(int x, int y) {
 28     for(int i = 0 ; i < 4 ; ++i) {
 29         int xx = x + tx[i], yy = y + ty[i];
 30         if(judge(xx, yy)) {
 31             G[x*m + y].push_back(xx*m + yy);
 32             du[x*m + y]++;
 33         }
 34     }
 35 }
 36 inline void change(int x, int y) {
 37     if(m == 1) {
 38         if(x < y) {
 39             str[x][0] = ‘^‘;
 40             str[y][0] = ‘v‘;
 41         } else {
 42             str[x][0] = ‘v‘;
 43             str[y][0] = ‘^‘;
 44         }
 45     } else {
 46         if(x - y == -m) {
 47             str[x/m][x%m] = ‘^‘;
 48             str[y/m][y%m] = ‘v‘;
 49         } else if(x - y == m) {
 50             str[x/m][x%m] = ‘v‘;
 51             str[y/m][y%m] = ‘^‘;
 52         } else if(x - y == -1) {
 53             str[x/m][x%m] = ‘<‘;
 54             str[y/m][y%m] = ‘>‘;
 55         } else {
 56             str[x/m][x%m] = ‘>‘;
 57             str[y/m][y%m] = ‘<‘;
 58         }
 59     }
 60 }
 61
 62 int main()
 63 {
 64     int cnt = 0, op = 0;
 65     scanf("%d %d", &n, &m);
 66     for(int i = 0; i < n; ++i) {
 67         scanf("%s", str[i]);
 68     }
 69     for(int i = 0; i < n; ++i) {
 70         for(int j = 0; j < m; ++j) {
 71             if(judge(i, j)) {
 72                 get(i, j);
 73                 ++op;
 74             }
 75         }
 76     }
 77     queue <int> que;
 78     for(int i = 0; i < n*m; ++i) {
 79         if(du[i] == 1)
 80             que.push(i);
 81     }
 82     while(!que.empty()) {
 83         int u = que.front();
 84         que.pop();
 85         du[u]--;
 86         for(int i = 0; i < G[u].size(); ++i) {
 87             int v = G[u][i];
 88             du[v]--;
 89             if(du[v] > 0) {
 90                 cnt++;
 91                 for(int j = 0; j < G[v].size(); ++j) {
 92                     du[G[v][j]]--;
 93                     if(du[G[v][j]] == 1) {
 94                         que.push(G[v][j]);
 95                     }
 96                 }
 97                 du[v] = 0;
 98                 change(u, v);
 99             } else if(du[v] == 0) {
100                 cnt++;
101                 change(u, v);
102             }
103         }
104     }
105     if(cnt * 2 == op) {
106         for(int i = 0; i < n; ++i) {
107             printf("%s\n", str[i]);
108         }
109     } else {
110         printf("Not unique\n");
111     }
112     return 0;
113 }