http://acm.hdu.edu.cn/showproblem.php?pid=5024
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Wang Xifeng's Little PlotTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description
《Dream of the Red Chamber》(also 《The Story of the Stone》) is one of
the Four Great Classical Novels of Chinese literature, and it is commonly regarded as the best one. This novel was created in Qing Dynasty, by Cao Xueqin. But the last 40 chapters of the original version is missing, and that part of current version was written by Gao E. There is a heart breaking story saying that after Cao Xueqin died, Cao's wife burned the last 40 chapter manuscript for heating because she was desperately poor. This story was proved a rumor a couple of days ago because someone found several pages of the original last 40 chapters written by Cao. In the novel, Wang Xifeng was in charge of Da Input
The map of Da Guan Yuan is represented by a matrix of characters '.'
and '#'. A '.' stands for a part of road, and a '#' stands for other things which one cannot step onto. When standing on a '.', one can go to adjacent '.'s through 8 directions: north, north-west, west, south-west, south, south-east,east and north-east. There are several test cases. For each case, the first line is an integer N(0<N<=100) ,meaning the map is a N × N matrix. Then the N × N matrix follows. The input ends with N = 0. Output
For each test case, print the maximum length of the road which Wang
Xifeng could find to locate Baoyu and Baochai's rooms. A road's length is the number of '.'s it includes. It's guaranteed that for any test case, the maximum length is at least 2. Sample Input
3
#.# ##. ..# 3 ... ##. ..# 3 ... ### ..# 3 ... ##. ... 0 Sample Output
3
4 3 5 Source
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hujie
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题意：
给出矩阵地图，.能走#不能走，八个方向都可以走，求某个点开始走一波直线然后转个90度的弯再走一波直线的最长的路能走多长。
题解：
预处理出每个点向各个方向能走多长，然后枚举拐弯处。
代码：
//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usll unsigned ll
#define mz(array) memset(array, 0, sizeof(array))
#define mf1(array) memset(array, -1, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) printf("%d\n",x);
#define RE freopen("D.in","r",stdin)
#define WE freopen("1biao.out","w",stdout)
#define mp make_pair
#define pb push_back
const double eps=1e-;
const double pi=acos(-1.0); const int gx[]= {,,-,};
const int gy[]= {,,,}; const int maxn=;
int n;
char a[maxn][maxn];
int b[maxn][maxn][];///0- 1| 2/ 3"\" b[x][y][dr]，在dr方向，以x,y为端点的最长线段的长度，包括x,y inline bool in(const int &x,const int &y) {
return(x>= && x<=n && y>= && y<=n);
} void gank(int X,int Y,int dr) {
int x=X,y=Y;
bool flag=;
int fx,fy;
int step=;
while(in(x,y)) {
//printf("a[%d][%d]=%c\n",x,y,a[x][y]);
if(!flag && a[x][y]=='.') {
flag=;
fx=x;
fy=y;
step=;
}
if(flag && a[x][y]=='#') {
flag=;
int nStep=;
//printf("%d,%d,%d,%d,%d\n",x,y,fx,fy,step);
while(in(fx,fy) && !(fx==x && fy==y)) {
b[fx][fy][dr]=max(nStep,step-nStep+);
fx+=gx[dr];
fy+=gy[dr];
nStep++;
}
}
x+=gx[dr];
y+=gy[dr];
step++;
} if(flag) {
flag=;
int nStep=;
//printf("%d,%d,%d,%d,%d\n",x,y,fx,fy,step);
while(in(fx,fy) && !(fx==x && fy==y)) {
b[fx][fy][dr]=max(nStep,step-nStep+);
fx+=gx[dr];
fy+=gy[dr];
nStep++;
}
} } int farm() {
int i,j;
mz(b); ///-
FOR(i,,n) {
gank(i,,);
} ///|
FOR(i,,n) {
gank(,i,);
} ///"/"
FOR(i,,n) {
gank(i,,);
//printf("start at %d,%d:\n",n,i);
gank(n,i,);
} ///"\"
FOR(i,,n) {
gank(i,,);
gank(,i,);
} int ans=;
FOR(i,,n) {
FOR(j,,n) {
//printf("%d,%d %d %d %d %d\n",i,j,b[i][j][0],b[i][j][1],b[i][j][2],b[i][j][3]);
ans=max(ans,b[i][j][]+b[i][j][]-);
ans=max(ans,b[i][j][]+b[i][j][]-);
}
}
return ans;
} int main() {
int i,j;
while(scanf("%d",&n)!=EOF) {
if(n==)break;
FOR(i,,n) {
FOR(j,,n)scanf(" %c",&a[i][j]);
} // printf("\n%d\n",n);
// FOR(i,1,n){
// FOR(j,1,n)printf("%c",a[i][j]);
// puts("");
// }
printf("%d\n",farm());
}
return ;
}