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

Inspired by the ice sculptures in Harbin, the members of the programming team from Arctic University of Robotics and Automata have decided to hold their own ice festival when they return home from the contest. They plan to harvest blocks of ice from a nearby lake when it freezes during the winter. To make it easier to monitor the thickness of the ice, they will lay out a rectangular grid over the surface of the lake and have a lightweight robot travel from square to square to measure ice thickness at each square in the grid. Three locations in the grid are specified as ``check-in" points and the robot is supposed to radio a progress report from these points when it is one-fourth, one-half, and three-fourths of the way through its tour of inspection. To avoid unnecessary wear and tear on the surface of the ice, the robot must begin its tour of the grid from the lower left corner, designated in (row,column) coordinates as (0,0), visiting every other grid location exactly once and ending its tour in row 0, column 1. Moreover, if there are multiple tours that the robot can follow, then a different one is to be used each day. The robot is able to move only one square per time step in one of the four compass directions: north, south, east, or west.

You are to design a program that determines how many different tours are possible for a given grid size and a sequence of three check-in points. For example, suppose the lake surface is marked off in a 3 x 6 grid and that the check-in points, in order of visitation, are (2,1), (2,4), and (0,4). Then the robot must start at (0,0) and end at (0,1) after visiting all 18 squares. It must visit location (2,1) on step 4 (= 18/4), location (2,4) on step 9 (= 18/2), and location (0,4) on step 13 (= 3 x 18/4). There are just two ways to do this (see Figure 8). Note that when the size of the grid is not divisible by 4, truncated division is used to determine the three check-in times.

Note that some configurations may not permit any valid tours at all. For example, in a 4 x 3 grid with check-in sequence (2,0), (3,2), and (0,2), there is no tour of the grid that begins at (0,0) and ends at (0,1).

Input 

The input contains several test cases. Each test case begins with a line containing two integers m and n, where 2m, n8, specifying the number of rows and columns, respectively, in the grid. This is followed by a line containing six integer values r₁, c₁, r₂, c₂, and r₃, c₃, where 0r_i < m and 0c_i < n for i = 1, 2, 3.

Following the last test case is a line containing two zeros.

Output 

Display the case number, beginning at 1, followed by the number of possible tours that begin at row 0, column 0, end at row 0, column 1, and visit row r_i, column c_i at time i x m x n/4 for i = 1, 2, 3. Follow the format of the sample output.

Sample Input 

3 6 
2 1 2 4 0 4 
4 3 
2 0 3 2 0 2 
0 0

Sample Output 

Case 1: 2 
Case 2: 0

---

題目描述：

給定起點終點，以及三個點要在指定步數恰好抵達，並且經過所有的點。
問有多少種走法。

題目解法：

剪枝方法非常多，有最短路內還不能走到該點，以及在指定步數前經過該點，又或者將圖形剖成兩半。

剖半的動作不能使用 BFS 去檢查，不然會太慢。

想法是看九宮格，把接下來要走的點放中間，查看是否分成兩個區域，看法是順時針，如果有兩個以上的區域被分開，即會產生 4 個以上的邊界(走過與沒走過)。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int n, m, r1, r2, r3, r4, c1,c2, c3, c4;
int step1, step2, step3, step4;
int used[10][10], ret;
int divid2(int x, int y) {
    static int v[9], i, j;
    v[0] = x-1 < 0 || y-1 < 0 || used[x-1][y-1];
    v[1] = x-1 < 0 || used[x-1][y];
    v[2] = x-1 < 0 || y+1 >= m || used[x-1][y+1];
    v[3] = y+1 >= m || used[x][y+1];
    v[4] = x+1 >= n || y+1 >= m || used[x+1][y+1];
    v[5] = x+1 >= n || used[x+1][y];
    v[6] = x+1 >= n || y-1 < 0 || used[x+1][y-1];
    v[7] = y-1 < 0 || used[x][y-1];
    v[8] = v[0];
    for(i = 0, j = 0; i < 8; i++)
        if(v[i] != v[i+1])
            j++;
    return j >= 4;
}
void dfs(int step, int x, int y) {
    if(step < step1 && used[r1][c1])   return;
    if(step < step2 && used[r2][c2])   return;
    if(step < step3 && used[r3][c3])   return;
    if(step < step4 && used[r4][c4])   return;
    if(step == step1 && (x != r1 || y != c1))
        return;
    if(step == step2 && (x != r2 || y != c2))
        return;
    if(step == step3 && (x != r3 || y != c3))
        return;
    if(step == step4 && (x != r4 || y != c4))
        return;
    if(step == step4) {ret++;return;}
    if(step < step1 && step+abs(x-r1)+abs(y-c1) > step1)
        return;
    if(step < step2 && step+abs(x-r2)+abs(y-c2) > step2)
        return;
    if(step < step3 && step+abs(x-r3)+abs(y-c3) > step3)
        return;
    if(step < step4 && step+abs(x-r4)+abs(y-c4) > step4)
        return;
    if(divid2(x, y))
        return;
    used[x][y] = 1;
    if(x+1 < n && !used[x+1][y])        dfs(step+1, x+1, y);
    if(x-1 >= 0 && !used[x-1][y])       dfs(step+1, x-1, y);
    if(y+1 < m && !used[x][y+1])        dfs(step+1, x, y+1);
    if(y-1 >= 0 && !used[x][y-1])       dfs(step+1, x, y-1);
    used[x][y] = 0;
}
int main() {
    int cases = 0;
    while(scanf("%d %d", &n, &m) == 2 && n+m) {
        scanf("%d %d", &r1, &c1), step1 = 1*n*m/4;
        scanf("%d %d", &r2, &c2), step2 = 2*n*m/4;
        scanf("%d %d", &r3, &c3), step3 = 3*n*m/4;
        r4 = 0, c4 = 1, step4 = n*m;
        memset(used, 0, sizeof(used));
        ret = 0;
        dfs(1, 0, 0);
        printf("Case %d: %d\n", ++cases, ret);
    }
    return 0;
}
/*
8 8
2 1 2 4 0 4
8 8
2 0 3 2 0 2
*/