洪水！（Flooded! ACM/ICPC World Finals 1999, UVa815）

有一个n*m（1≤m，n<30）的网格，每个格子是边长10米的正方形，网格四周是无限大
的墙壁。输入每个格子的海拔高度，以及网格内雨水的总体积，输出水位的海拔高度以及有
多少百分比的区域有水（即高度严格小于水平面）

解题思路非常直观可以直接看代码

#include <bits/stdc++.h>


#define mem(a,b) memset(a,b,sizeof(a))
typedef long long int LL;
const int maxn = 1005;
const int inf = 0x3f3f3f3f;#include <bits/stdc++.h>


#define mem(a,b) memset(a,b,sizeof(a))
typedef long long int LL;
const int maxn = 1005;
const int inf = 0x3f3f3f3f;
using namespace std;

//int vaule[maxn];
int main(void)
{
    #ifdef MYLOCAKYACM
//    freopen("input.in","r",stdin);
//   	freopen("output.out","w",stdout);
    #endif // MYLOCA
	int n,m;
	int v[maxn];
	int kase = 0;
	while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
	{
		mem(v,0);
		for(int i = 0;i<n*m;i++)
		{
			scanf("%d",&v[i]);
		}
		sort(v,v+m*n);
		v[m*n] = inf;//保证在最后一定会break; 
		double maxd;scanf("%lf",&maxd);
		maxd /= 100;
		int i = 0;
		for(;i<n*m;i++)
		{
			maxd += v[i];
			if(maxd/(i+1)<=v[i+1])break;
		}
		printf("Region %d\n",++kase);
		printf("Water level is %.2lf meters.\n",(double)maxd/(i+1));
		printf("%.2lf percent of the region is under water.\n\n",(i+1)*100.0/(m*n));
	} 
	return 0;
}


using namespace std;

//int vaule[maxn];
int main(void)
{
    #ifdef MYLOCAKYACM
//    freopen("input.in","r",stdin);
//   	freopen("output.out","w",stdout);
    #endif // MYLOCA
	int n,m;
	int v[maxn];
	int kase = 0;
	while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
	{
		mem(v,0);
		for(int i = 0;i<n*m;i++)
		{
			scanf("%d",&v[i]);
		}
		sort(v,v+m*n);
		v[m*n] = inf;
		double maxd;scanf("%lf",&maxd);
		maxd /= 100;
		int i = 0;
		for(;i<n*m;i++)
		{
			maxd += v[i];
			if(maxd/(i+1)<=v[i+1])break;
		}
		printf("Region %d\n",++kase);
		printf("Water level is %.2lf meters.\n",(double)maxd/(i+1));
		printf("%.2lf percent of the region is under water.\n\n",(i+1)*100.0/(m*n));
	} 
	return 0;
}