Create 5725_summer_via.cpp

From http://blog.csdn.net/summer_via/article/details/51979297
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Kirigaya Kazuto 2017-07-21 20:31:27 +08:00 committed by GitHub
parent 5e9f7d6b13
commit 7d3fcd085a

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HDOJ/5725_summer_via.cpp Normal file
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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
#include<queue>
#include<vector>
#include<list>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
#define INF 0x3f3f3f3f
int n,m;
long long dp[1005][1005];
long long dp2[505][505];
char mp[1005][1005];
//n*m的图计算以11为起点到所有曼哈顿距离和
long long get(long long n,long long m)
{
return m*(m-1)*n/2+n*(n-1)*m/2;
}
//n*m的图计算任意点到所有点的曼哈顿距离和
long long get2(long long x,long long y)
{
return get(x,y)+get(n-x+1,y)+get(x,m-y+1)+get(n-x+1,m-y+1)-(get(1,x)+get(1,y)+get(1,n-x+1)+get(1,m-y+1));
}
int xx[1005],yy[1005],cnt2;
int pr[1005],pc[1005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cnt2=0;
scanf("%d%d",&n,&m);
long long up=(long long)m*n*(m*n-1)*(m+n)/3,down=0;
int cnt=0;
memset(pr,-1,sizeof pr);
memset(pc,-1,sizeof pc);
getchar();
for(int i=0;i<n;i++)
{
gets(mp[i]);
for(int j=0;j<m;j++)
{
if(mp[i][j]=='G')
{
pr[i]=j;
pc[j]=i;
up-=4ll*j*(m-j-1);//同行被阻断会在后面重复计算
up-=4ll*i*(n-i-1);//同列类似
up-=2*get2(i+1,j+1),cnt++;
xx[cnt2]=i;
yy[cnt2]=j;
cnt2++;
}
}
}
//计算绕行路径和
int suml=0,sumr=0;
for(int i=0;i<n;i++)
{
if(pr[i]!=-1)
{
if(i==0||pr[i]>pr[i-1]) suml+=pr[i];
else suml=pr[i];
up+=4ll*suml*(m-pr[i]-1);
}
else suml=0;
if(pr[i]!=-1)
{
if(i==0||pr[i]<pr[i-1]) sumr+=m-pr[i]-1;
else sumr=m-pr[i]-1;
up+=4ll*sumr*pr[i];
}
else sumr=0;
}
int sumu=0,sumd=0;
for(int i=0;i<m;i++)
{
if(pc[i]!=-1)
{
if(i==0||pc[i]>pc[i-1]) sumu+=pc[i];
else sumu=pc[i];
up+=4ll*sumu*(n-pc[i]-1);
}
else sumu=0;
if(
pc[i]!=-1)
{
if(i==0||pc[i]<pc[i-1]) sumd+=n-pc[i]-1;
else sumd=n-pc[i]-1;
up+=4ll*sumd*pc[i];
}
else sumd=0;
}
for(int i=0;i<cnt2;i++)
{
for(int j=0;j<cnt2;j++)
{
up+=abs(xx[i]-xx[j])+abs(yy[i]-yy[j]);
}
}
down=(long long)(n*m-cnt)*(n*m-cnt);
double ans=(long double)up/down;
printf("%.4llf\n",ans);
}
return 0;
}