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HDU 5465——Clarke and puzzle——————【树状数组BIT维护前缀和+Nim博弈】

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Clarke and puzzle

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 223


Problem Description
Clarke is a patient with multiple personality disorder. One day, Clarke split into two personality a and b, they are playing a game. 
There is a nm matrix, each grid of this matrix has a number ci,j
a wants to beat b every time, so a ask you for a help. 
There are q operations, each of them is belonging to one of the following two types: 
1. They play the game on a (x1,y1)(x2,y2) sub matrix. They take turns operating. On any turn, the player can choose a grid which has a positive integer from the sub matrix and decrease it by a positive integer which less than or equal this grid‘s number. The player who can‘t operate is loser. a always operate first, he wants to know if he can win this game. 
2. Change ci,j to b

 

 

Input
The first line contains a integer T(1T5), the number of test cases. 
For each test case: 
The first line contains three integers n,m,q(1n,m500,1q2105) 
Then nm matrix follow, the i row j column is a integer ci,j(0ci,j109) 
Then q lines follow, the first number is opt
if opt=1, then 4 integers x1,y1,x1,y2(1x1x2n,1y1y2m) follow, represent operation 1
if opt=2, then 3 integers i,j,b follow, represent operation 2.
 

 

Output
For each testcase, for each operation 1, print Yes if a can win this game, otherwise print No.
 

 

Sample Input
1
1 2 3
1 2
1 1 1 1 2
2 1 2 1
1 1 1 1 2
 

 

Sample Output
Yes
No
 
 
Hint: The first enquiry: $a$ can decrease grid $(1, 2)$‘s number by $1$. No matter what $b$ operate next, there is always one grid with number $1$ remaining . So, $a$ wins. The second enquiry: No matter what $a$ operate, there is always one grid with number $1$ remaining. So, $b$ wins.
 

 

Source
 
 
题目大意:给你t组测试数据。每组测试数据中有一组n,m,q,分别表示有一个n*m的矩阵,有q组询问。每组询问中,操作1表示查询(x1,y1) ---(x2,y2)在矩阵中做Nim游戏先手是否会赢。操作2表示要把矩阵中某个值修改为c。
 
 
知识补充:对于一维树状数组见的比较多。二维还是见的少(弱)。那么二维跟一维的区别是:一般意义上二维用来求的是矩阵的和。对于C[x][y],这里的定义就是从(1,1)---(x,y)矩阵的元素和。我们可以将每行抽象成一个点,这样其实就是抽象成了求一维的情况。其实一维跟二维实现的写法差别很容易想到。对于二维BIT,我们得到C数组,需要n^2log(n)^2的时间复杂度。
  
求和:
int sum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)){
        for(int j=y;j>0;j-=lowbit(j)){
            ret+=C[i][j];
        }
    }
    return ret;
}

修改:

void modify(int x,int y,int val){
    for(int i=x;i<=n;i+=lowbit(i)){
        for(int j=y;j<=m;j+=lowbit(j)){
            C[i][j] += val;
        }
    }
}

  

 
 
解题思路:技术分享
 
 
#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
const int maxn=550;
int a[maxn][maxn];
int C[maxn][maxn];
int n,m;
int lowbit(int x){
    return x & (-x);
}
void modify(int x,int y,int val){
    for(int i=x;i<=n;i+=lowbit(i)){
        for(int j=y;j<=m;j+=lowbit(j)){
            C[i][j] ^= val;
        }
    }
}
int sum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)){
        for(int j=y;j>0;j-=lowbit(j)){
            ret ^=C[i][j];
        }
    }
    return ret;
}
int main(){
    int t,q;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        memset(C,0,sizeof(C));
        scanf("%d%d%d",&n,&m,&q);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                modify(i,j,a[i][j]);
            }
        }
        int typ,x1,x2,y1,y2,c;
        for(int i=0;i<q;i++){
            scanf("%d",&typ);
            if(typ==1){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                int ans1,ans2,ans3,ans4;
                ans1=sum(x2,y2);
                ans2=sum(x2,y1-1);
                ans3=sum(x1-1,y2);
                ans4=sum(x1-1,y1-1);
                int ans=ans1^ans2^ans3^ans4;
                printf("%s\n",ans==0?"No":"Yes");
            }else{
                scanf("%d%d%d",&x1,&y1,&c);
                modify(x1,y1,c^a[x1][y1]);
                a[x1][y1]=c;
            }
        }
    }
    return 0;
}

  

 
 

HDU 5465——Clarke and puzzle——————【树状数组BIT维护前缀和+Nim博弈】

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原文地址:http://www.cnblogs.com/chengsheng/p/4832894.html

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