POJ2030The Secret Number题解动态规划DP

The Secret Number

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1682		Accepted: 554

Description

Your job is to find out the secret number hidden in a matrix, each of whose element is a digit ('0'-'9') or a letter ('A'-'Z'). You can see an example matrix in Figure 1.

Figure 1: A Matrix
The secret number and other non-secret ones are coded in a matrix as sequences of digits in a decimal format. You should only consider sequences of digits D1 D2 ... Dn such that Dk+1 (1 <= k < n) is either right next to or immediately below Dk in the matrix. The secret you are seeking is the largest number coded in this manner.

Four coded numbers in the matrix in Figure 1, i.e., 908820, 23140037, 23900037, and 9930, are depicted in Figure 2. As you may see, in general, two or more coded numbers may share a common subsequence. In this case, the secret number is 23900037, which is the largest among the set of all coded numbers in the matrix.

Figure 2: Coded Numbers
In contrast, the sequences illustrated in Figure 3 should be excluded: 908A2 includes a letter; the fifth digit of 23149930 is above the fourth; the third digit of 90037 is below right of the second.

Figure 3: Inappropriate Sequences
Write a program to figure out the secret number from a given matrix.

Input

The input consists of multiple data sets, each data set representing a matrix. The format of each data set is as follows.

W H
C11C12 ... C1W
C21C22 ... C2W
...
CH1CH2 ... CHW

In the first line of a data set, two positive integers W and H are given. W indicates the width (the number of columns) of the matrix, and H indicates the height (the number of rows) of the matrix. W+H is less than or equal to 70.

H lines follow the first line, each of which corresponds to a row of the matrix in top to bottom order. The i-th row consists of W characters Ci1Ci2 ... CiW in left to right order. You may assume that the matrix includes at least one non-zero digit.

Following the last data set, two zeros in a line indicate the end of the input.

Output

For each data set, print the secret number on a line. Leading zeros should be suppressed.

Sample Input

7 4
9R2A993
0E314A0
8A900DE
820R037
6 7
JH03HE
ID7722
0DA1AH
30C9G5
99971A
CA7EAI
AHLBEM
20 2
A1234567891234CBDEGH
BDEDF908034265091499
0 0

Sample Output

23900037
771971
12345908034265091499

Source

Japan 2003 Domestic

经典的取数模型，关键是记录大数

刚开始用string数组超时，改为结构体32ms

状态：

d[i][j]表示到第i行j列的最大Secret Number

状态转移方程：

d[i][j]=max(d[i-1][j],d[i][j-1])+mat[i][j]

代码：

#include<cstdio> #include<cctype> #include<iostream> #include<string> using namespace std; struct node { char s[71]; node() {memset(s,0,sizeof(s));} bool operator<(const node t)const { if(strlen(s)==strlen(t.s)) return strcmp(s,t.s)<0; return strlen(s)<strlen(t.s); } void operator+=(char c) { s[strlen(s)]=c; } }; int main() { int m,n,i,j; char map[105][105]; while(scanf("%d%d",&m,&n),m+n) { node d[105][105],ans; for(i=1;i<=n;i++) scanf("%s",map[i]+1); for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(isdigit(map[i][j])) { if(d[i-1][j]<d[i][j-1]) d[i][j]=d[i][j-1]; else d[i][j]=d[i-1][j]; if(map[i][j]!='0'||strlen(d[i][j].s)) d[i][j]+=map[i][j]; if(ans<d[i][j]) ans=d[i][j]; } printf("%s/n",ans.s); } }