D. Drazil and Tiles
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Drazil created a following problem about putting1×2tiles into ann×mgrid:
"There is a grid with some cells that are empty and some cells that are occupied. You should use1×2tiles to cover all empty cells and no two tiles should cover each other. And you should print a solution about how to do it."
But Drazil doesn’t like to write special checking program for this task. His friend, Varda advised him: "how about asking contestant only to print the solutionwhen it exists and it is unique? Otherwise contestant may print ‘Not unique’ ".
Drazil found that the constraints for this task may be much larger than for the original task!
Can you solve this new problem?
Note that you should print ‘Not unique’ either when there exists no solution or when there exists several different solutions for the original task.
Input
The first line contains two integersnandm(1≤n,m≤2000).
The followingnlines describe the grid rows. Character ‘.’ denotes an empty cell, and the character ‘*’ denotes a cell that is occupied.
Output
If there is no solution or the solution is not unique, you should print the string "Not unique".
Otherwise you should print how to cover all empty cells with1×2tiles. Use characters "<>" to denote horizontal tiles and characters "^v" to denote vertical tiles. Refer to the sample test for the output format example.
Sample test(s)
input
3 3….*….
output
Not unique
input
4 4..***…*.**….
output
<>***^<>*v**<><>
input
2 4*..*….
output
*<>*<><>
input
1 1.
output
Not unique
input
1 1*
output
*
Note
In the first case, there are indeed two solutions:
<>^^*vv<>
and
^<>v*^<>v
so the answer is "Not unique".
思路：用有点类似拓扑排序的方法做。。一直找度为1的结点，直到找不到为止，，这里的度的意思是’.’挨着的’.’的个数
注意这里由于数据输入输出比较大，最好用scanf和printf来输入输出，直接用cin和cout会超时，如果硬要用cin和cout可以加一句
ios::sync_with_stdio(false);
AC代码：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;const int maxn = 2005;char a[maxn][maxn];pair<int ,int> p[maxn * maxn];int mox[4] = {-1, 0, 1, 0};int moy[4] = {0, 1, 0, -1};char cg[4] = {'v', '<', '^', '>'};int n, m;bool ispoint(int x, int y) {return (x >= 0 && x < n && y >= 0 && y < m && a[x][y] == '.');}int deg(int x, int y) {int ans = 0;for(int i = 0; i < 4; i++) {int fx = x + mox[i];int fy = y + moy[i];if(ispoint(fx, fy)) ans++;}return ans;}int main() {while(scanf("%d %d", &n, &m) != EOF) {for(int i = 0; i < n; i++)scanf("%s", a[i]);int h = 0, sum = 0;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)if(ispoint(i, j) && deg(i, j) == 1)p[sum++] = make_pair(i, j);for(; h < sum; h++) {int x = p[h].first, y = p[h].second;for(int i = 0; i < 4; i++) {int fx = x + mox[i];int fy = y + moy[i];if(ispoint(fx, fy)) {a[x][y] = cg[i];a[fx][fy] = cg[i ^ 2];for(int i = 0; i < 4; i++) {int ffx = fx + mox[i];int ffy = fy + moy[i];if(ispoint(ffx, ffy) && deg(ffx, ffy) == 1)p[sum++] = make_pair(ffx, ffy);}break;}}}int flag = 0;for(int i = 0; i < n && !flag; i++)for(int j = 0; j < m && !flag; j++)if(a[i][j] == '.' ) flag = 1;if(flag == 1) printf("Not unique\n");else {for(int i = 0; i < n; i++, printf("\n"))for(int j = 0; j < m; j++)printf("%c", a[i][j]);} }return 0;}
让所有的愁向后飞去。请不要回头去追你因该向前奔跑，因为快乐在前方!
