UVA 815 Flooded!

题意：来自：https://blog.csdn.net/lecholin/article/details/70186673

思路：

①数组存每个网格的高度，然后排序，做题时想象为上面的柱状图。

②注意对淹没的处理。

③注意每个输出都有空行。

 #include "stdio.h"
 #include "stdlib.h"
 #include "string.h"
 #define FIN freopen("input.txt","r",stdin)
 #define FOUT freopen("output.txt","w",stdout)
 int main()
 {
     //FIN;
     //FOUT;
     int m, n;
     int data[],kase=;
     double level, percentage, high = ,sum;
     while (scanf("%d%d",&m,&n)==&&n!=&&m!=)
     {
         kase++;
         for (int i = ; i < m*n; i++)
         {
             scanf("%d", &data[i]);
         }
         high = ;
         scanf("%lf", &sum);
         sum /= ;//正方形为10*10，化简一下sum
         //排序
         for (int i = ; i < m*n; i++)
         {
             for (int j = i + ; j < m*n; j++)
             {
                 if (data[i] > data[j])
                 {
                     int ex = data[i];
                     data[i] = data[j];
                     data[j] = ex;
                 }
             }
         }
         //data[m*n] = 100000;//最大的边界
         //输出
         /*printf("排序后:\n");
         for (int i = 0; i < m*n; i++)
         {
             printf("%-4d", data[i]);
             if ((i + 1) % m == 0)
                 printf("\n");
         }*/
         for (int i = ; i < n*m; i++)
         {
             if (sum != - && i == m*n - )//找到最后一个说明淹没了
             {
                 high += (sum*1.0) / (i + );
                 high += data[];
                 int cntDown = (i + );
                 percentage = (cntDown*1.0) / (m*n) * ;
                 sum = -;
                 break;
             }
             if ((data[i + ] - data[i])*(i+)< sum)//小就继续往上面走
             {
                 sum -= (data[i + ] - data[i])*(i+);
                 high += data[i+]-data[i];
             }
             else
             {
                 high += (sum*1.0) / (i+);
                 high += data[];
                 int cntDown =(i+);
                 percentage = (cntDown*1.0) / (m*n)*;
                 sum = -;
                 break;
             }

         }
         printf("Region %d\n", kase);
         printf("Water level is %.2f meters.\n",high);
         printf("%.2f percent of the region is under water.\n\n",percentage);
     }
     return ;
 }