Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Tom has developed a special kind of puzzle: it involves a whole bunch of identical puzzle pieces. The pieces have the shape of three adjoint squares in an L-shape. The corner square is black, the two adjacent squares are white.

The puzzler is given a pattern of black and white squares in a rectangular grid. The challenge is to create that pattern using these pieces. The pieces can be rotated, but must not overlap.

Tom has already designed a few nice patterns, but he needs to find out if they can be constructed with the pieces at all. Rather than trying to test this for each pattern by hand, he wants to write a computer program to determine this for him. Can you help him?

Input 

On the first line a positive integer: the number of test cases, at most 100. After that per test case:

• one line with two integers n and m (1n, m500): the height and width of the grid containing the pattern, respectively.
• n lines, each containing m characters, denoting the grid. Each character is `B', `W', or `.', indicating a black, white or empty square respectively.

The grid contains at least one black or white square.

Output 

Per test case:

• one line with either ``YES" or ``NO", indicating whether or not it is possible to construct the pattern with the puzzle pieces. You may assume that there is an infinite supply of pieces.

Sample Input 

2
3 4
BWW.
WWBW
..WB
3 3
W..
BW.
WBW

Sample Output 

YES
NO

---

題目描述：

用上述 L 型拼圖進行完美覆蓋。可能輸出 "YES"，反之 "NO"。

題目解法：

自己常用的 maxflow 太慢，可能要換其他 maxflow 的演算法。
不過這題匈牙利算法在底線上通過了。

將每一個 B 拆成 2 個點，分別為水平跟垂直。
也就是說 B 只會從水平匹配到一個 W，垂直匹配到一個 W。
因此問題就是要詢問拆點後，能不能進行完美匹配。

非常底線的做法，接近 TLE orz

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <queue>
using namespace std;
struct Node {
    int y;
    int next;
} edge[1000005];
int e, head[250005];
void addEdge(int x, int y) {
    edge[e].y = y;
    edge[e].next = head[x], head[x] = e++;
}
int mx[250005], my[250005], used[250005];
int dfs(int now) {
    int i, x;
    for(i = head[now]; i != -1; i = edge[i].next) {
        x = edge[i].y;
        if(!used[x]) {
            used[x] = 1;
            if(my[x] == -1 || dfs(my[x])) {
                mx[now] = x, my[x] = now;
                return 1;
            }
        }
    }
    return 0;
}
int main() {
    int testcase, n, m;
    int i, j;
    scanf("%d", &testcase);
    char g[505][505];
    int node[505][505];
    while(testcase--) {
        scanf("%d %d", &n, &m);
        for(i = 0; i < n; i++)
            scanf("%s", g[i]);
        e = 0;
        memset(head, -1, sizeof(head));
        int B = 0, W = 0;
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                if(g[i][j] == 'W')
                    node[i][j] = W++;
                if(g[i][j] == 'B')
                    node[i][j] = B++;
            }
        }
        if(W != B*2) {
            puts("NO");
            continue;
        }
        int source = 0, sink = W + B*2 + 1;
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                if(g[i][j] == 'B') {
                    // horizontal
                    if(j-1 >= 0 && g[i][j-1] == 'W')
                        addEdge(node[i][j], node[i][j-1]);
                    if(j+1 < m && g[i][j+1] == 'W')
                        addEdge(node[i][j], node[i][j+1]);
                    // vertical
                    if(i-1 >= 0 && g[i-1][j] == 'W')
                        addEdge(node[i][j]+B, node[i-1][j]);
                    if(i+1 < n && g[i+1][j] == 'W')
                        addEdge(node[i][j]+B, node[i+1][j]);
                }
            }
        }
        memset(mx, -1, sizeof(mx));
        memset(my, -1, sizeof(my));
        int match = 0;
        B = B*2;
        for(i = 0; i < B; i++) {
            if(mx[i] == -1) {
                memset(used, 0, sizeof(used));
                if(dfs(i))
                    match++;
                else
                    break;// cut condition.
            }
        }
        puts(match == B ? "YES" : "NO");
    }
    return 0;
}