标签：模拟

对每个环都进行遍历，找出最大值

如果值相等的话就可以还要判断下走的步数

Rotation Lock Puzzle

Alice was felling into a cave. She found a strange door with a number square matrix. These numbers can be rotated around the center clockwise or counterclockwise. A fairy came and told her how to solve this puzzle lock: “When the sum of main diagonal and anti-diagonal is maximum, the door is open.”.
Here, main diagonal is the diagonal runs from the top left corner to the bottom right corner, and anti-diagonal runs from the top right to the bottom left corner. The size of square matrix is always odd.



This sample is a square matrix with 5*5. The numbers with vertical shadow can be rotated around center ‘3’, the numbers with horizontal shadow is another queue. Alice found that if she rotated vertical shadow number with one step, the sum of two diagonals is maximum value of 72 (the center number is counted only once).

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
0

72 1

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define MAXN 12

int num[12][12];

int main(){
    int n,i,j,k;

    while(~scanf("%d",&n)){
        if(n == 0){
            break;
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&num[i][j]);
            }
        }
        int ans = 0;
        int anscon = 0;
        for(k=1;k<=n/2;k++){
            int marksum = -1;
            int mark = k;
            int sum=0;
            for(i=k;i<=n-k;i++){
                sum = 0;
                //printf("%d %d %d %d\n",num[k][i],num[i][n-k+1],num[n-k+1][n-i+1],num[n-i+1][k]);
                sum+=num[k][i];
                sum+=num[i][n-k+1];
                sum+=num[n-k+1][n-i+1];
                sum+=num[n-i+1][k];
                //printf("%d\n",sum);
                if(sum > marksum){
                    marksum = sum;
                    mark = min(i-k,(n-k+1-i));
                }
                if(sum == marksum){
                    int mark1 = min(i-k,n-k+1-i);
                    mark = min(mark1,mark);
                }
            }
            //printf("%d\n",num[n/2+1][n/2+1]);
            ans+=marksum;
            anscon+=mark;
        }
        printf("%d %d\n",ans+num[n/2+1][n/2+1],anscon);
    }

    return 0;
}

HDU 4708

标签：模拟

原文地址：http://blog.csdn.net/zcr_7/article/details/41450633