Cow Hopscotch

题目描述

Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a variant of the game for themselves to play. Being played by clumsy animals weighing nearly a ton, Cow Hopscotch almost always ends in disaster, but this has surprisingly not deterred the cows from attempting to play nearly every afternoon.

The game is played on an R by C grid (2 <= R <= 750, 2 <= C <= 750), where each square is labeled with an integer in the range 1..K (1 <= K <= R*C). Cows start in the top-left square and move to the bottom-right square by a sequence of jumps, where a jump is valid if and only if

1) You are jumping to a square labeled with a different integer than your current square,

2) The square that you are jumping to is at least one row below the current square that you are on, and

3) The square that you are jumping to is at least one column to the right of the current square that you are on.

Please help the cows compute the number of different possible sequences of valid jumps that will take them from the top-left square to the bottom-right square.

输入

The first line contains the integers R, C, and K. The next R lines will each contain C integers, each in the range 1..K.

输出

Output the number of different ways one can jump from the top-left square to the bottom-right square, mod 1000000007.

样例输入

4 4 4
1 1 1 1
1 3 2 1
1 2 4 1
1 1 1 1

样例输出

5
分析:2个难点,状态转移和分治处理;
   状态转移:ans[x][y]=∑ans[i][j](i<x,j<y)-Σans[i][j](i<x,j<y,a[i][j]=a[x][y]);
   分治:由于处理左半部分,上半部分时和右半部分,下半部分没有半毛钱关系,所以分治处理;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e3+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,a[maxn][maxn],ans[maxn][maxn],check[maxn*maxn];
void gao(int l,int r)
{
if(l==r)return;
int mid=l+r>>;
gao(l,mid);
memset(check,,sizeof(check));
int all=;
for(int j=;j<=m;j++)
{
for(int i=mid+;i<=r;i++)
ans[i][j]=((ans[i][j]+all-check[a[i][j]])%mod+mod)%mod;
for(int i=l;i<=mid;i++)
all+=ans[i][j],all%=mod,check[a[i][j]]+=ans[i][j],check[a[i][j]]%=mod;
}
gao(mid+,r);
}
int main()
{
int i,j;
scanf("%d%d%d",&n,&m,&k);
rep(i,,n)rep(j,,m)scanf("%d",&a[i][j]);
ans[][]=;
gao(,n);
printf("%d\n",ans[n][m]);
//system("pause");
return ;
}

Cow Hopscotch的更多相关文章

  1. 【BZOJ3939】[Usaco2015 Feb]Cow Hopscotch 动态规划+线段树

    [BZOJ3939][Usaco2015 Feb]Cow Hopscotch Description Just like humans enjoy playing the game of Hopsco ...

  2. 【洛谷】2990:[USACO10OPEN]牛跳房子Cow Hopscotch【单调队列优化DP】

    P2990 [USACO10OPEN]牛跳房子Cow Hopscotch 题目描述 The cows have reverted to their childhood and are playing ...

  3. 洛谷P3120 [USACO15FEB]Cow Hopscotch

    题目描述 Just like humans enjoy playing the game of Hopscotch, Farmer John&apos;s cows have invented ...

  4. 【bzoj3939】[Usaco2015 Feb]Cow Hopscotch 动态开点线段树优化dp

    题目描述 Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a varian ...

  5. 洛谷 P3120 [USACO15FEB]牛跳房子(金)Cow Hopscotch (Gold)

    P3120 [USACO15FEB]牛跳房子(金)Cow Hopscotch (Gold) 就像人类喜欢跳格子游戏一样,FJ的奶牛们发明了一种新的跳格子游戏.虽然这种接近一吨的笨拙的动物玩跳格子游戏几 ...

  6. Cow Hopscotch (单调队列 + DP)

    链接:https://ac.nowcoder.com/acm/contest/1113/K来源:牛客网 The cows have reverted to their childhood and ar ...

  7. BZOJ3939 : [Usaco2015 Feb]Cow Hopscotch

    设f[i][j]表示到(i,j)的方案数,则有 $f[i][j]=\sum f[x][y](x<i,y<j,a[x][y]!=a[i][j])=\sum f[x][y](x<i,y& ...

  8. bzoj3939 【USACO 2015 FEB GOLD 】cow hopscotch

    Description 就像人类喜欢玩"跳房子"的游戏,农民约翰的奶牛已经发明了该游戏的一个变种自己玩.由于笨拙的动物体重近一吨打,牛跳房子几乎总是以灾难告终,但这是没有阻止奶牛几 ...

  9. BZOJ1915[USACO 2010 Open Gold 1.Cow Hopscotch]——DP+斜率优化

    题目描述 奶牛们正在回味童年,玩一个类似跳格子的游戏,在这个游戏里,奶牛们在草地上画了一行N个格子,(3 <=N <= 250,000),编号为1..N.就像任何一个好游戏一样,这样的跳格 ...

随机推荐

  1. ios 控件 UIButton

    - (void)setTitle:(NSString *)title forState:(UIControlState)state; //设置标题 - (void)setTitleColor:(UIC ...

  2. linux服务器无法显示tomcat详细错误信息

    严重: One or more listeners failed to start. Full details will be found in the appropriate container l ...

  3. python--windows下安装BeautifulSoup

    python有很多内置的模块可以不安装使用,用起来非常方便,但是也有一些挺有用的非内置的模块不能直接使用,需要话费点力气手动安装. 进入python安装目录下的Scripts目录,查看是否有pip工具 ...

  4. Asp.Net调用Office组件操作时的DCOM配置 (转)

    Asp.Net调用Office组件操作时的DCOM配置 http://blog.csdn.net/gz775/article/details/6447758 在项目中将数据导出为Excel格式时出现“ ...

  5. iOS 最值宏定义

    NSIntegerMax.NSIntegerMin.NSUIntegerMax.MAXFLOAT

  6. POJ 1308/并查集

    题目链接 /* 判断一棵树: * 1.There is exactly one node, called the root, to which no directed edges point. * 2 ...

  7. ReactiveCocoa的冷信号与热信号 探讨

    背景 ReactiveCocoa(简称RAC)是最初由GitHub团队开发的一套基于Cocoa的FRP框架.FRP即Functional Reactive Programming(函数式响应式编程), ...

  8. html5 localStorage和sessionStorage

    该随便来自: http://www.cnblogs.com/yuzhongwusan/archive/2011/12/19/2293347.html html5中的Web Storage包括了两种存储 ...

  9. android sql Cursor

    Cursor 是每行的集合. 使用 moveToFirst() 定位第一行. 你必须知道每一列的名称.你必须知道每一列的数据类型.Cursor 是一个随机的数据源. 所有的数据都是通过下标取得. Cu ...

  10. 优化のzencart URL &zenid=.....

    zencart URL后面带有一串&zenid=.....解决方案 发布时间:2013年3月16日 次浏览:106 经木木测试,此方法可用. ================= 最近一个客户的 ...