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 + 5;
int n, m, tx[] = {-1, 0, 1, 0}, ty[] = {0, -1, 0, 1};
char str[N][N];
int du[N*N];
vector <int> G[N*N];
inline bool judge(int x, int y) {
    if(x >= 0 && x < n && y >= 0 && y < m && str[x][y] == '.')
        return true;
    return false;
}
inline void get(int x, int y) {
    for(int i = 0 ; i < 4 ; ++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 == 1) {
        if(x < y) {
            str[x][0] = '^';
            str[y][0] = 'v';
        } else {
            str[x][0] = 'v';
            str[y][0] = '^';
        }
    } 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 == -1) {
            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 = 0, op = 0;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; ++i) {
        scanf("%s", str[i]);
    }
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < m; ++j) {
            if(judge(i, j)) {
                get(i, j);
                ++op;
            }
        }
    }
    queue <int> que;
    for(int i = 0; i < n*m; ++i) {
        if(du[i] == 1)
            que.push(i);
    }
    while(!que.empty()) {
        int u = que.front();
        que.pop();
        du[u]--;
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i];
            du[v]--;
            if(du[v] > 0) {
                cnt++;
                for(int j = 0; j < G[v].size(); ++j) {
                    du[G[v][j]]--;
                    if(du[G[v][j]] == 1) {
                        que.push(G[v][j]);
                    }
                }
                du[v] = 0;
                change(u, v);
            } else if(du[v] == 0) {
                cnt++;
                change(u, v);
            }
        }
    }
    if(cnt * 2 == op) {
        for(int i = 0; i < n; ++i) {
            printf("%s
", str[i]);
        }
    } else {
        printf("Not unique
");
    }
    return 0;
}