poj1877模拟题

  一道world final的水题。说起这道题还有个小故事，我在msra当 intern时的舍友，曾经去面google总部inern时候被问的一道题目和这道题目挺像的，只是从二维变成一维，后来一个同学提起，就顺便刷了下。

  最大的难点是理解题目，其实就是顺着题目的意思对每块地的海拔排个序后，一块地一块地的遍历过来就好，不过注意各种极端情况。

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
int regionIndex=0;
int n,m;
double region[1000];
double totalWaterCubic;
double waterLevel;
double floodedPercent;

scanf("%d%d",&m,&n);

while(n!=0 && m!=0)
{
regionIndex++;

for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
scanf("%lf",®ion[i*m+j]);
}
scanf("%lf",&totalWaterCubic);

if(n==1&&m==1)
{
if(totalWaterCubic>0.0f)
{
waterLevel=totalWaterCubic/100.0f + region[0];
floodedPercent=100.00f;
}
else
{
waterLevel=0.0f;
floodedPercent=0.0f;
}
}

else
{
if(totalWaterCubic==0.0f)
{
waterLevel=0.0f;
floodedPercent=0.0f;
}

else
{
sort(region,region+n*m);

double curFloodedCubic=0.0f;
int curFloodedIndex=0;
for(int i=1;i<n*m;i++)
{
//If square i is flooded, it need addedFloodedCubic cubic of water.
double addedFloodedCubic=(region[i]-region[i-1])*i*100.0f;
if(curFloodedCubic+addedFloodedCubic<totalWaterCubic)//the square is flooded
{
curFloodedCubic+=addedFloodedCubic;

if(i==n*m-1)//All the squares are flooded.
{
double leftWaterCubic=totalWaterCubic-curFloodedCubic;
waterLevel=leftWaterCubic/((i+1)*100.0f)+region[i];
floodedPercent=100.00f;
break;
}
}
else//Those squares whose height are higher than square i-1 are no more flooded.
{
double leftWaterCubic= totalWaterCubic-curFloodedCubic;
waterLevel=leftWaterCubic/(i*100.0f)+region[i-1];
floodedPercent=(double)i/(double)(n*m)* 100.0f;
break;
}
}
}
}

//Outpu the result
printf("Region %d\n",regionIndex);
printf("Water level is %.2lf meters.\n",waterLevel);
printf("%.2lf percent of the region is under water.\n",floodedPercent);
printf("\n");

scanf("%d%d",&m,&n);
}

return 0;
}