HDU 2125 Local area network

简单DP，N×M的网格其中有一条边坏掉了，问从起点到终点的放法数

有两种方法，一种是DP很好理解

 

 1 //#define LOCAL

 2 #include <cstdio>

 3 #include <cstring>

 4 

 5 int dp[42][42];

 6 bool flag[42][42];

 7 

 8 int main(void)

 9 {

10     #ifdef LOCAL

11         freopen("2125in.txt", "r", stdin);

12     #endif

13 

14     int r, c;

15     while(scanf("%d%d", &r, &c) == 2)

16     {

17         int y1, x1, y2, x2;

18         scanf("%d%d%d%d", &y1, &x1, &y2, &x2);

19         dp[0][1] = 1;

20         memset(flag, false, sizeof(flag));

21         flag[x1+1][y1+1] = flag[x2+1][y2+1] = true;

22         for(int i = 1; i <= r; ++i)

23             for(int j = 1; j <= c; ++j)

24             {

25                 dp[i][j] = dp[i-1][j] + dp[i][j-1];

26                 if(flag[i][j])

27                 {

28                     if(flag[i][j-1])

29                         dp[i][j] -= dp[i][j-1];

30                     if(flag[i-1][j])

31                         dp[i][j] -= dp[i-1][j];

32                 }

33             }

34         printf("%d\n", dp[r][c]);

35     }

36     return 0;

37 }

代码君

 

第二种，用数学公式

如果没有坏边的话，总放法数是C^N-1_(M+N-2) 

因为每种方法都要走(M+N-2)步，向上走N-1步，向下走M-1步

现在考虑一条坏边，那么就计算经过这条坏边的方案数然后从总数里面减去即可

经过坏边的放法数就是从起点到(x1, y1)的放法数×从(x2, y2)到终点的放法数

 1 #define LOCAL

 2 #include <algorithm>

 3 #include <cstdio>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 long long c(long long m, long long n)

 8 {

 9     if(m==0 || n==0)

10         return 1;

11     n = min(n, m-n);

12     long long ans = 1;

13     for(int i = 0; i < n; ++i)

14         ans = ans * (m-i) / (1+i);

15     return ans;

16 }

17 

18 int main(void)

19 {

20     #ifdef LOCAL

21         freopen("2125in.txt", "r", stdin);

22     #endif

23 

24     int n, m;

25     while(scanf("%d%d", &n, &m) == 2)

26     {

27         int x1, y1, x2, y2;

28         scanf("%d%d%d%d", &x1, &y1, &x2, &y2);

29         if(x1+y1 > x2+y2)

30         {

31             int t = x1;

32             x1 = x2, x2 = t;

33             t = y1;

34             y1 = y2, y2 = t;

35         }

36         printf("%lld\n", c(m+n-2, m-1) - c(x1+y1, x1) * c(m+n-2-x2-y2, m-1-x2));

37     }

38     return 0;

39 }

代码君