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HDU - 1505 City Game(dp)

题意:求由R和F组成的矩阵中,最大的由F组成的矩阵面积。

分析:

1、hdu1506http://www.cnblogs.com/tyty-Somnuspoppy/p/7341431.html与此题相似。

2、将矩阵的每行看成1506中的坐标轴分别处理。

3、算出以该行为坐标轴,坐标轴上,每个由F组成的矩形的高度,与1506的区别是,1506中每个矩形高度已知。

4、按照1506的方法分别算出每个矩形最左边和最右边连续比其高的矩形的下标,通过(r[i][j] - l[i][j] + 1) * h[i][j]比较即可。

5、h[i][j]就相当于1506中a[i]。

h[i][j]----以第i行为坐标轴,第j个矩形的高度。

a[i]-----在坐标轴上,第i个矩形的高度。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 1000 + 10;const int MAXT = 10000 + 10;using namespace std;int h[MAXN][MAXN], l[MAXN][MAXN], r[MAXN][MAXN];int main(){    int K;    scanf("%d", &K);    while(K--){        memset(h, 0, sizeof h);        memset(l, 0, sizeof l);        memset(r, 0, sizeof r);        int m, n;        char c[5];        scanf("%d%d", &m, &n);        for(int i = 1; i <= m; ++i){            for(int j = 1; j <= n; ++j){                scanf("%s", c);                if(c[0] == ‘F‘) h[i][j] = h[i - 1][j] + 1;                else h[i][j] = 0;            }        }        int ans = 0;        for(int i = 1; i <= m; ++i){            l[i][1] = 1;            for(int j = 2; j <= n; ++j){                int tmp = j;                while(tmp > 1 && h[i][tmp - 1] >= h[i][j]) tmp = l[i][tmp - 1];                l[i][j] = tmp;            }            r[i][n] = n;            for(int j = n - 1; j >= 1; --j){                int tmp = j;                while(tmp < n && h[i][tmp + 1] >= h[i][j]) tmp = r[i][tmp + 1];                r[i][j] = tmp;            }            for(int j = 1; j <= n; ++j){                ans = max(ans, (r[i][j] - l[i][j] + 1) * h[i][j]);            }        }        printf("%d\n", ans * 3);    }    return 0;}

  

HDU - 1505 City Game(dp)