1.题目描述：点击打开链接
2.解题思路：本题一开始迟迟没有好的思路，随后通过思考，发现应该用贪心法解决：首先应该着力填充周围只有一个空格的点，，随后以它的空格为中心，再不断向外拓展，直到整个界面被填充。这样便解决了本题。
3.代码：
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 2005char a[N][N];typedef pair<int, int>P;queue<P>q;int dx[] = { -1, 0, 1, 0 };//dx,dy必须和cg一一对应int dy[] = { 0, 1, 0, -1 };char cg[] = "v<^>";int n, m;bool is_point(int x, int y){return (x >= 0 && x < n&&y >= 0 && y < m&&a[x][y] == '.');}int deg(int x, int y)//找(x,y)周围空格的个数，称为度数{int ans = 0;for (int i = 0; i < 4; i++){int nx = x + dx[i];int ny = y + dy[i];if (is_point(nx, ny))ans++;}return ans;}int main(){//freopen("test.txt", "r", stdin);while (scanf("%d%d", &n, &m) != EOF){for (int i = 0; i < n; i++)scanf("%s", a[i]);for (int i = 0; i < n;i++)for (int j = 0; j < m; j++)if (is_point(i, j) && deg(i, j) == 1)//将度为1的点入队列q.push(P(i, j));while (!q.empty())//利用BFS向外填充，由于退出循环时队列为空，因此不必清空队列{P u = q.front(); q.pop();int x = u.first, y = u.second;for (int k = 0; k < 4; k++){int nx = x + dx[k];int ny = y + dy[k];if (is_point(nx, ny)){a[x][y] = cg[k];a[nx][ny] = cg[k ^ 2];for (int l = 0; l < 4; l++){int mx = nx + dx[l];int my = ny + dy[l];if (is_point(mx, my) && deg(mx, my) == 1)q.push(P(mx, my));}}}}int flag = 0;for (int i = 0; i < n; i++)//查找是否还有点没有被填充{for (int j = 0; j < m; j++)if (a[i][j] == '.'){flag = 1;break;}if (flag)break;}if (flag)printf("Not unique\n");else{for (int i = 0; i < n; i++, printf("\n"))for (int j = 0; j < m; j++)putchar(a[i][j]);}}return 0;}
发光并非太阳的专利，你也可以发光
