Problem DescriptionThere is a n×m board, a chess want to go to the position
(n,m) from the position (1,1).
The chess is able to go to position (x2,y2) from the position (x1,y1), only and if only x1,y1,x2,y2 is satisfied that (x2−x1)2+(y2−y1)2=5, x2>x1, y2>y1.
Unfortunately, there are some obstacles on the board. And the chess never can stay on the grid where has a obstacle.
I want you to tell me, There are how may ways the chess can achieve its goal.InputThe input consists of multiple test cases.
For each test case:
The first line is three integers, n,m,r,(1≤n,m≤1018,0≤r≤100), denoting the height of the board, the weight of the board, and the number of the obstacles on the board.
Then follow r lines, each lines have two integers, x,y(1≤x≤n,1≤y≤m), denoting the position of the obstacles. please note there aren't never a obstacles at position (1,1).OutputFor each test case,output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer after module 110119.Sample Input1 1 03 3 04 4 12 14 4 13 27 10 21 27 1Sample Output
Case #1: 1Case #2: 0Case #3: 2Case #4: 1Case #5: 5
一个n*m的棋盘,一个棋子想从(1,1)走到(n,n)
棋子每步可从(x1,y1)到达(x2,y2),当且仅当
另外,棋盘上有r处障碍,棋子无法到达障碍处
问棋子从(1,1)到达(n,m)有多少种走法
【类型】
lucas定理求组合数+dp
【分析】
首先我们要确定棋子是怎么走的由式子
,x2>x1,y2>y1可知
方程有两组解,x2=x1+2,y2=y1+1或x2=x1+1,y2=y1+2
即当棋子处于位置(x,y)时,它可以一步到达①(x+1,y+2)或②(x+2,y+1)处
那我们不妨求解一下通式,假设从位置(x1,y1)到(x2,y2)需要①走法走k1步,②走法走k2步
那么可得方程组
求解得
然后,因为从(x1,y1)到(x2,y2)在无障碍的情况下与走法先后无关,只与k1,k2有关
所以从(x1,y1)到(x2,y2)的方法数为
或
组合数?还是大组合数,显然,lucas派上用场了
那么,障碍怎么办?比赛的时候想到了容斥,但感觉若r个障碍都可达的话,容斥复杂度还是蛮高的,但是官方题解给的就是容斥
算了,我们还是管自己的方法来解,dp
令dp[i]表示到第i个障碍前面不经过任何障碍的方案数然后转移的时候算出(1,1)到第i个障碍的方案数枚举j从1到i-1所有的障碍,减去从(1,1)出发只经由障碍j再到障碍i的方案数最后将重点(n,m)作为第r+1个障碍
ans就是dp[r+1]
这样复杂度是O(r^2)的【时间复杂度&&优化】
O(r^2)
View Code1 #include<algorithm> 2 3 #include<iostream> 4 #include<stdio.h> 5 #include<map> 6 #include<vector> 7 #include<math.h> 8 #include<string.h> 9 using namespace std; 10 #define ll long long 11 #define mod 110119 12 #define INF 0x3f3f3f3f 13 #define pai (sqrt(5)+ 1)/2 14 #define pi acos(-1.0) 15 ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} //最大公约数 16 ll lcm(ll a,ll b){return a/gcd(a,b)*b;}//最小公倍数 17 ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} 18 //求平方数 19 double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;} 20 ll f[mod+10]; 21 ll dp[102]; 22 struct aa 23 { 24 ll x; 25 ll y; 26 }a[102]; 27 bool cmp(aa t1,aa t2) 28 { 29 if(t1.x==t2.x)return t1.y<t2.y; //排序 30 return t1.x<t2.x; 31 } 32 void init() 33 { 34 f[0]=1; 35 for(int i=1;i<=mod;i++)//初始化阶乘 36 f[i]=(f[i-1]*i)%mod; 37 } 38 ll C(ll n,ll m) 39 { 40 if(n<m)return 0; 41 return f[n]*powmod(f[m]*f[n-m]%mod,mod-2,mod);//费马定理 42 } 43 ll lucas(ll n,ll m) 44 { 45 if(m==0)return 1; 46 return C(n%mod,m%mod)*lucas(n/mod,m/mod)%mod;//大组合数 47 } 48 bool judge(int i,int j) 49 { 50 if(a[i].x<a[j].x||a[i].y<a[j].y) 51 return false; 52 return true; 53 } 54 ll cal(ll i,ll j)//是否可以从i到达j 55 { 56 ll t1=a[j].x-a[i].x,t2=a[j].y-a[i].y; 57 ll k1=2*t2-t1,k2=2*t1-t2; 58 if(k1<0||k2<0||k1%3||k2%3) 59 return 0; 60 k1/=3; 61 k2/=3; 62 return lucas(k1+k2,k1); 63 } 64 int main() 65 { 66 init(); 67 int case1=1; 68 ll n,m,r; 69 while(~scanf("%lld %lld %lld",&n,&m,&r)) 70 { 71 printf("Case #%d: ",case1++); 72 for(int i=1;i<=r;i++) 73 scanf("%lld %lld",&a[i].x,&a[i].y); 74 a[0].x=a[0].y=1; 75 a[r+1].x=n; 76 a[r+1].y=m; 77 sort(a+1,a+1+r,cmp); 78 for(int i=1;i<=r+1;i++) 79 { 80 dp[i]=cal(0,i); 81 for(int j=1;j<i;j++) 82 if(judge(i,j)) 83 dp[i]=(dp[i]+mod-cal(j,i)*dp[j]%mod)%mod; 84 } 85 printf("%I64d ",dp[r+1]); 86 } 87 return 0; 88 }
