【codeforces 515D】Drazil and Tiles

【题目链接】:http://codeforces.com/contest/515/problem/D

【题意】

给你一个n*m的格子;
然后让你用1*2的长方形去填格子的空缺;
如果有填满的方案且方案是唯一的;
则输出那个方案,否则,输出不唯一;

【题解】

记录每个点的度;
每个点的度,为这个点4个方向上空格的个数;
优先处理度数为1的点;
这些点的摆放方式肯定是唯一的;
摆完这些点(两个之后),与之相连的点的度数都减1；
看看有没有新的度数为1的点;
很像拓扑排序对吧。
最后看看占据的点是不是n*m个

【Number Of WA】

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define ps push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 2e3+100;

struct abc
{
    int x, y;
};

int n, m,du[N][N],cnt = 0;
char s[N][N];
bool bo[N][N];
queue <abc> dl;

void cl(int x, int y)
{
    rep1(j, 1, 4)
    {
        int x1 = x + dx[j], y1 = y + dy[j];
        du[x1][y1]--;
        if (bo[x1][y1] && du[x1][y1] == 1)
            dl.push({ x1,y1 });
    }
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    memset(bo, false, sizeof bo);
    rei(n), rei(m);
    rep1(i, 1, n)
    {
        scanf("%s", s[i] + 1);
        rep1(j, 1, m)
            if (s[i][j] == '.')
                bo[i][j] = true;
            else
                bo[i][j] = false,cnt++;
    }
    rep1(i,1,n)
        rep1(j,1,m)
        if (bo[i][j])
        {
            int num = 0;
            rep1(k, 1, 4)
                num += bo[i + dx[k]][j + dy[k]];
            du[i][j] = num;
            if (du[i][j] == 1)
            {
                dl.push({ i,j });
            }
        }
    while (!dl.empty())
    {
        int x0 = dl.front().x, y0 = dl.front().y;
        dl.pop();
        rep1(j, 1, 4)
        {
            int x = x0 + dx[j], y = y0 + dy[j];
            /*
            const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
            const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
            */
            if (bo[x][y])
            {
                if (j == 1) s[x0][y0] = '^', s[x][y] = 'v';
                if (j == 2) s[x0][y0] = 'v', s[x][y] = '^';
                if (j == 3)s[x0][y0] = '>', s[x][y] = '<';
                if (j == 4) s[x0][y0] = '<', s[x][y] = '>';
                bo[x][y] = bo[x0][y0] = false;
                cl(x, y), cl(x0, y0);
                cnt += 2;
                break;
            }
        }
    }
    if (cnt == n*m)
    {
        rep1(i, 1, n)
            puts(s[i] + 1);
    }
    else
        puts("Not unique");
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}