ZOJ1074 最大和子矩陣 DP最大連續子序列
To the Max
Time Limit:1 Second Memory Limit: 32768 KB
Problem
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the
sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines).
These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Example
Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Output
15
最多100行100列,對每個舉行都要考慮到2500*2500,
當然計算每個矩形的元素和之前要預處理
,
令每行的[j,k}]素之和為a區域,s[i][j][k]表示從第一行到第i行的a區域之和,
所有對於行從[i,k],列從[p,q]的矩形,其元素和=s[k][p][q]-s[i-1][p][q];
#include<cstdio> #include<string> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> #include<climits> #include<queue> #include<vector> #include<map> #include<sstream> #include<set> #include<stack> #include<utility> //#pragma comment(linker, "/STACK:102400000,102400000") #define INF 0x3f3f3f3f #define PI 3.1415926535897932384626 #define eps 1e-10 #define sqr(x) ((x)*(x)) #define FOR0(i,n) for(int i=0 ;i<(n) ;i++) #define FOR1(i,n) for(int i=1 ;i<=(n) ;i++) #define FORD(i,n) for(int i=(n) ;i>=0 ;i--) #define lson num<<1,l,mid #define rson num<<1|1,mid+1,r #define MID int mid=(l+r)>>1 #define zero(x)((x>0? x:-x)<1e-15) using namespace std; const int maxn= 100+10 ; //const int maxm= ; //const int INF= ; //typedef long long ll; //ifstream fin("input.txt"); //ofstream fout("output.txt"); //fin.close(); //fout.close(); //freopen("a.in","r",stdin); //freopen("a.out","w",stdout); int a[maxn][maxn]; int s[maxn][maxn][maxn]; int row[maxn][maxn]; inline int ReadInt() { int flag=0; char ch = getchar(); int data = 0; while (ch < '0' || ch > '9') { if(ch=='-') flag=1; ch = getchar(); } do { data = data*10 + ch-'0'; ch = getchar(); }while (ch >= '0' && ch <= '9'); if(flag) data=-data; return data; } int main() { int i,j;int n; while(~scanf("%d",&n)) { memset(row,0,sizeof row); FOR1(i,n) { FOR1(j,n) { a[i][j]=ReadInt(); row[i][j]=row[i][j-1]+a[i][j]; } } memset(s,0,sizeof s); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { for(int k=j+1;k<=n;k++) { s[i][j][k]=s[i-1][j][k]+row[i][k]-row[i][j-1]; } } } int maxi=-INF; for(int i=1;i<=n;i++) { for(int k=i+1;k<=n;k++) { for(int p=1;p<=n;p++) { for(int q=p+1;q<=n;q++) { maxi=max(maxi,s[k][p][q]-s[i-1][p][q]); } } } } printf("%d\n",maxi); } return 0; } /* 4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2 */
後來發現,更好的辦法是利用最大連續子序列,對第從第i行到第j行的矩形,其長度用最大連續子序列來求
#include<cstdio> #include<string> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> #include<climits> #include<queue> #include<vector> #include<map> #include<sstream> #include<set> #include<stack> #include<utility> //#pragma comment(linker, "/STACK:102400000,102400000") #define INF 0x3f3f3f3f #define PI 3.1415926535897932384626 #define eps 1e-10 #define sqr(x) ((x)*(x)) #define FOR0(i,n) for(int i=0 ;i<(n) ;i++) #define FOR1(i,n) for(int i=1 ;i<=(n) ;i++) #define FORD(i,n) for(int i=(n) ;i>=0 ;i--) #define lson num<<1,l,mid #define rson num<<1|1,mid+1,r #define MID int mid=(l+r)>>1 #define zero(x)((x>0? x:-x)<1e-15) using namespace std; const int maxn= 100+20 ; //const int maxm= ; //const int INF= ; //typedef long long ll; //ifstream fin("input.txt"); //ofstream fout("output.txt"); //fin.close(); //fout.close(); //freopen("a.in","r",stdin); //freopen("a.out","w",stdout); int n; int col[maxn][maxn]; int a[maxn][maxn]; int main() { int i,j; int n; while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) { col[i][0]=0; } FOR1(i,n) FOR1(j,n) { scanf("%d",&a[i][j]); col[j][i]=col[j][i-1]+a[i][j]; } int maxi=-INF; for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int sum=0; for(int k=1;k<=n;k++) { int t=col[k][j]-col[k][i-1]; if(sum+t>=t) sum+=t; else sum=t; maxi=max(sum,maxi); } } } printf("%d\n",maxi); } return 0; }