标签：style blog http color os io for ar art
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 }
第二种，用数学公式
如果没有坏边的话，总放法数是CN-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 }
标签：style blog http color os io for ar art
原文地址：http://www.cnblogs.com/AOQNRMGYXLMV/p/3931672.html
