La 3029 City Game (scanning method)

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Idea: Process Each lattice: calculates the largest rectangle that contains it.

UP[I][J]: The height of the storage rectangle;

LEFT[I][J]: The column number that stores the left edge of the rectangle;

RIGHT[I][J]: Stores the column number of the right boundary of the rectangle.

Area = up[i][j] * (Right[i][j]-left[i][j] +1).

Experience: Even with ideas, write code before you draw on paper, simulation. Otherwise, like the code in the notice of the place, written in 0, it is estimated that debug half a day do not know where the wrong.

#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include < string> #include <stack> #include <queue> #include <set> #include <map>typedef long long ll; using namespace std;const int inf=0x3f3f3f3f;const int maxn=1e3+10;int mat[maxn][maxn],up[maxn][maxn];int MYLEFT[MAXN] [Maxn],myright[maxn][maxn];int M,n;int Main () {int t;scanf ("%d", &t), while (t--) {scanf ("%d%d", &m,&n); char ch;for (int i=0;i<m;i++) {for (int j=0;j<n;j++) {Ch=getchar (), while (ch!= ' R ' && ch!= ' F ') Ch=getchar (); if (ch== ' R ') Mat[i][j]=1;else mat[i][j]=0;}} /*for (int i=0;i<m;i++) {for (int j=0;j<n;j++) {printf ("%d", Mat[i][j]);} printf ("\ n");} */int ans=0,i,j;for (i=0;i<m;i++) {//from top to bottom scan int Lo=-1,ro=n;//lo represents the column number of the left obstacle, ro the same. for (j=0;j<n;j++) {//left-to-right scanning, processing up[][] and left[][]. if (mat[i][j]==1) {lo=j;up[i][j]=0;myleft [i][j]=0;} Else{up[i][j]= (i==0?1:up[i-1][j]+1); myleft[i][j]= (I==0?lo+1:max (myleft[i-1][j],lo+1));//!!!}} for (int j=n-1;j>=0;j--) {//right-to-left scanning, processing right[][]. if (mat[i][j]==1) {ro=j;myright[i][j]=n;//attention to detail}else{myright[i][j]= (I==0?ro-1:min (myright[i-1][j],ro-1));//!!! Ans=max (ans,up[i][j]* (myright[i][j]-myleft[i][j]+1));}} printf ("%d\n", ans*3);} return 0;}

　　

La 3029 City Game (scanning method)