UVA - 815 Flooded!

To enable homebuyers to estimate the cost of flood insurance, a real-estate firm provides clients with
the elevation of each 10-meter by 10-meter square of land in regions where homes may be purchased.
Water from rain, melting snow, and burst water mains will collect first in those squares with the
lowest elevations, since water from squares of higher elevation will run downhill. For simplicity, we also
assume that storm sewers enable water from high-elevation squares in valleys (completely enclosed by
still higher elevation squares) to drain to lower elevation squares, and that water will not be absorbed
by the land.
From weather data archives, we know the typical volume of water that collects in a region. As
prospective homebuyers, we wish to know the elevation of the water after it has collected in lowlying
squares, and also the percentage of the region’s area that is completely submerged (that is, the
percentage of 10-meter squares whose elevation is strictly less than the water level). You are to write
the program that provides these results.
Input
The input consists of a sequence of region descriptions. Each begins with a pair of integers, m and
n, each less than 30, giving the dimensions of the rectangular region in 10-meter units. Immediately
following are m lines of n integers giving the elevations of the squares in row-major order. Elevations
are given in meters, with positive and negative numbers representing elevations above and below sea
level, respectively. The final value in each region description is an integer that indicates the number of
cubic meters of water that will collect in the region. A pair of zeroes follows the description of the last
region.
Output
For each region, display the region number (1, 2, . . . ), the water level (in meters above or below sea
level) and the percentage of the region’s area under water, each on a separate line. The water level
and percentage of the region’s area under water are to be displayed accurate to two fractional digits.
Follow the output for each region with a blank line.
Sample Input
3 3
25 37 45
51 12 34
94 83 27
10000
0 0
Sample Output
Region 1
Water level is 46.67 meters.

66.67 percent of the region is under water.

先按照高度从小到大的顺序对所有方格进行排序。然后一步一步模拟水淹没一个又一个方格的过程。当水量超过给定量的时候，跳出循环。如果是由于方格全部淹没导致跳出循环的需要另外处理。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;double a[1000];//转换成一维数组int main(){double now,n,m,water;int kase=0;while(cin>>m>>n){     int i;    if(m==0&&n==0)        break;   for( i=1;i<=m*n;i++)    cin>>a[i];   cin>>water;   water/=100;//除以每个方格的面积   now=0;   double high,percent;   int sum=m*n;    sort(a+1,a+sum+1);    a[0]=a[1];//为了计算水淹没的高度，以最低的方格为参考平面    for(i=1;i<=m*n;i++)        a[i]-=a[0];    for( i=2;i<=m*n;i++)        {now+=(a[i]-a[i-1])*(i-1);//依次模拟水淹没方格        if(now>water) break;        }        if(i!=m*n+1)        {            now-=(a[i]-a[i-1])*(i-1);            high=a[i-1]+(water-now)/(i-1)+a[0];//多余的水量无法淹没整块方格。计算淹没的高度 percent=(i-1)/(m*n)*100;        }        else{            high=a[sum]+(water-now)/(m*n)+a[0];//淹没完了所有方格，计算在这之后还会淹没多高            percent=100;        }        printf("Region %d\n",++kase);printf("Water level is %.2f meters.\n",high);printf("%.2f percent of the region is under water.\n\n",percent);}return 0;}