UVA1330 City Game

链接

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4076

题解

悬线法，每个点记录上方离他最近的障碍点，把该点和当前点的连线成为悬线，记录左右平移悬线碰到障碍前能够到达的最远的点，利用这些信息找到最大矩形。

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 1010
#define clear(x) memset(x,0,sizeof(x))
using namespace std;
int mat[maxn][maxn], n, m, left[maxn][maxn], right[maxn][maxn], up[maxn][maxn];
char readc()
{
    char c;
    for(c=getchar();c^'F' and c^'R';c=getchar());
    return c;
}
void init()
{
    int i, j;
    clear(mat), clear(left), clear(right), clear(up);
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)mat[i][j]=readc()=='F'?0:1;
    for(i=1;i<=n;i++)mat[i][0]=mat[i][m+1]=1;
    for(j=1;j<=m;j++)mat[0][j]=mat[n+1][j]=1, up[0][j]=0;
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)up[i][j]=mat[i][j]?i:up[i-1][j];
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)left[i][j]=mat[i][j]?j:left[i][j-1];
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(!mat[i-1][j])left[i][j]=max(left[i][j],left[i-1][j]);
    for(i=1;i<=n;i++)right[i][m+1]=m+1;
    for(i=1;i<=n;i++)for(j=m;j;j--)right[i][j]=mat[i][j]?j:right[i][j+1];
    for(i=1;i<=n;i++)for(j=m;j;j--)if(!mat[i-1][j])right[i][j]=min(right[i][j],right[i-1][j]);
}
int main()
{
    int i, j, ans, T;
    for(scanf("%d",&T);T--;)
    {
        init();
        ans=-1; 
        for(i=1;i<=n;i++)for(j=1;j<=m;j++)ans=max(ans,(right[i][j]-left[i][j]-1)*(i-up[i][j]));
        printf("%d\n",ans*3);
    }
    return 0;
}