Largest Rectangle in a Histogram(dp)
http://acm.hdu.edu.cn/showproblem.php?pid=1506
题意：给出n个矩形的高度，每个矩形的宽都为1，求相邻的矩形能组合成的最大的矩形的面积。
思路：求出比第i个矩形大的最左边的矩形的位置 l[i], 及比第i个矩形大的最右边的矩形的位置 r[i], 则第i个矩形的面积 s = (r[i]-l[i]+1)*hign[i]。
如果第i-1个矩形比第i个矩形大，则 l[i] 必定在 l[i-1]的 左边，同理，r[i]必定在 r[i+1]的右边。
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #define LL __int64 5 const int N=100002; 6 using namespace std; 7 LL high[N],l[N],r[N]; 8 int main() 9 { 10 int n; 11 while(~scanf("%d",&n)&&n) 12 { 13 for (int i = 1; i <= n; i++) 14 scanf("%I64d",&high[i]); 15 for (int i = 1; i <= n; i++) 16 { 17 int pos = i; 18 while(pos > 1 && high[pos-1] >= high[i]) 19 { 20 pos = l[pos-1];//如果第i个矩形左边的矩形(pos-1)比i高， 21 //则pos移动到l[pos-1](即比第pos-1个矩形高的最左边的矩形的位置) 22 } 23 l[i] = pos;//l[i]表示比第i个矩形连续高的最左边的矩形位置 24 } 25 for (int i = n; i >= 1; --i) 26 { 27 int pos = i; 28 while(pos < n && high[i] <= high[pos+1]) 29 { 30 pos = r[pos+1]; 31 } 32 r[i] = pos;//表示比第i个矩形连续高的最右边的矩形位置 33 } 34 LL maxn = 0; 35 for (int i = 1; i <= n; i++) 36 { 37 maxn = max((r[i]-l[i]+1)*high[i],maxn); 38 } 39 printf("%I64d\n",maxn); 40 } 41 return 0; 42 }
同类型的题：
http://acm.hdu.edu.cn/showproblem.php?pid=1505
1 #include <stdio.h> 2 #include <algorithm> 3 #include <string.h> 4 #define LL __int64 5 const int N=100002; 6 using namespace std; 7 LL l[N],r[N],high[N]; 8 9 LL max_area(int m) 10 { 11 for (int i = 1; i <= m; i++) 12 { 13 if(i > 1 && high[i] <= high[i-1]) 14 { 15 l[i] = l[i-1]; 16 } 17 else 18 l[i] = i; 19 } 20 for (int i = m; i >= 1; i--) 21 { 22 23 if(i < m && high[i] <= high[i+1]) 24 { 25 r[i] = r[i+1]; 26 } 27 else 28 r[i] = i; 29 } 30 LL maxn = 0; 31 for (int i = 1; i <= m; i++) 32 { 33 maxn = max((r[i]-l[i]+1)*high[i],maxn); 34 } 35 return maxn; 36 } 37 int main() 38 { 39 int t; 40 char s[23]; 41 scanf("%d",&t); 42 while(t--) 43 { 44 int n,m; 45 LL ans = 0; 46 scanf("%d %d",&n,&m); 47 memset(high,0,sizeof(high)); 48 for (int i = 1; i <= n; i++) 49 { 50 for (int j = 1; j <= m; j++) 51 { 52 scanf("%s",s); 53 if (s[0]=='F') 54 high[j]++; 55 else 56 high[j] = 0; 57 } 58 ans = max(max_area(m),ans);//第一行到第i行最大的面积 59 } 60 printf("%I64d\n",ans*3); 61 } 62 return 0; 63 }
