HDU 2125 Local Area Network

A simple dp,nxm grid where one edge is broken, ask for the number of points from the starting point to the end point

There are two methods, one is DP very well understood

1 //#define LOCAL2#include <cstdio>3#include <cstring>4 5 intdp[ the][ the];6 BOOLflag[ the][ the];7 8 intMainvoid)9 {Ten #ifdef LOCAL OneFreopen ("2125in.txt","R", stdin); A     #endif -  -     intR, C; the      while(SCANF ("%d%d", &r, &c) = =2) -     { -         inty1, x1, y2, x2; -scanf"%d%d%d%d", &y1, &x1, &y2, &x2); +dp[0][1] =1; -memset (Flag,false,sizeof(flag)); +flag[x1+1][y1+1] = flag[x2+1][y2+1] =true; A          for(inti =1; I <= R; ++i) at              for(intj =1; J <= C; ++j) -             { -DP[I][J] = dp[i-1][J] + dp[i][j-1]; -                 if(Flag[i][j]) -                 { -                     if(flag[i][j-1]) inDP[I][J]-= dp[i][j-1]; -                     if(flag[i-1][j]) toDP[I][J]-= dp[i-1][j]; +                 } -             } theprintf"%d\n", Dp[r][c]); *     } $     return 0;Panax Notoginseng}

code June

The second, using a mathematical formula

If there is no bad side, the total number of methods is CN-1 (m+n-2)

Because each method has to walk (m+n-2) step, walk up N-1 step, walk down M-1 step

Now consider a bad side, then calculate the number of schemes passing through this bad side and subtract from the total

The number of X1 by the bad side is the number of Y1 from the starting point to the (x2, y2) to the end of the method.

1 #defineLOCAL2#include <algorithm>3#include <cstdio>4#include <cstring>5 using namespacestd;6 7 Long LongCLong LongMLong LongN)8 {9     if(m==0|| n==0)Ten         return 1; Onen = min (n, M-n); A     Long LongAns =1; -      for(inti =0; I < n; ++i) -Ans = ans * (m-i)/(1+i); the     returnans; - } -  - intMainvoid) + { - #ifdef LOCAL +Freopen ("2125in.txt","R", stdin); A     #endif at  -     intN, M; -      while(SCANF ("%d%d", &n, &m) = =2) -     { -         intx1, y1, x2, y2; -scanf"%d%d%d%d", &x1, &y1, &AMP;X2, &y2); in         if(X1+y1 > x2+y2) -         { to             intt =X1; +x1 = x2, x2 =T; -t =Y1; they1 = y2, y2 =T; *         } $printf"%lld\n", C (m+n-2, M-1)-C (x1+y1, x1) * C (m+n-2-x2-y2, M1-x2));Panax Notoginseng     } -     return 0; the}

code June

HDU 2125 Local Area Network