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 intusing 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;}/*59 3 2 5 97 4 7 5 46 9 3 9 35 2 8 7 29 9 4 1 9*/