HDU 4708 Rotation Lock Puzzle && 2013 ACM/ICPC Asia Regional Online —— Warmup

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/11391275

题意：该矩阵中，可以顺时针或逆时针旋转每一圈，问旋转至少几次可以使主对角线和副对角线和最大

下面代码先计算每一圈的和最大，最后加上中间的点

题目保证矩阵边长为奇数

 

做起来有点繁琐，下面k表示第k圈，最中间是第0圈，只要for那一圈的第一排数字即可。

 

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cctype>
#include <queue>
#include <stdlib.h>
#include <cstdlib>
#include <set>
#include <math.h>
#include <vector>
#define inf 107374182
#define N 105
#define im -1000000
#define ll int
using namespace std;
inline ll Max(ll a,ll b){return a>b?a:b;}
inline ll Min(ll a,ll b){return a<b?a:b;}
int map[N][N],n;
int sum(int k,int h,int l){

    k=(n/2)+1-k;
    return map[k+h][k+l];
}
int sum1(int k,int h,int l){
    int len=k*2;
    k=(n/2)+1-k;
    int x=h,y=l;
    if(x>len){y+=(x-len);x=len;}
    if(x<0){y-=0-x;x=0;}

    if(y>len){x+=(y-len);y=len;}
    if(y<0){x-=0-y;y=0;}

    return map[k+x][k+y];
}
int GO(int x,int y,int k){
    if(x==0)return Min(y,k-y);
    return Min(x,k-x);
}
int abs(int k,int i){
    if(i>k)return i-2*(i-k);
    return i+2*(k-i); 
}
int main(){
    ll i,j,k,go;
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&map[i][j]);
        int step=0,ans=0;
        int zhongdian=(n/2)+1;
        for(k=1;k<=(n>>1);k++)
        {
            int len=k*2,maxx=-1,temp;
            go=10000;
            for(i=0;i<k*2+1;i++)
            {
                temp=sum(k,0,i)+sum(k,len,abs(k,i));
                temp+= sum1(k,0,i+len)+sum1(k,len,abs(k,i)-len);
                if(temp>maxx)
                {
                    maxx=temp;
                    go=GO(0,i,len);
                }
                else if(temp==maxx)
                    if(go>GO(0,i,len))
                        go=GO(0,i,len);

            }
            step+=go;
            ans+=maxx;
        }
        printf("%d %d\n",ans+map[zhongdian][zhongdian],step);
    }
    return 0;
}
/*

5
9 3 2 5 9
7 4 7 5 4
6 9 3 9 3
5 2 8 7 2
9 9 4 1 9

*/