【集训笔记】动态规划【HDOJ1159【HDOJ1003

终于开始DP了】

HDOJ_1159  Common  Subsequence

题目链接

Sample Input
abcfbc abfcab
programming contest
abcd mnp

Sample Output
  4
  2
  0

子结构特征：

 

lf(i,j)= 　　  f(i-1,j-1)+1 (a[i]==b[j])　

　　　　　　max(f(i-1,j),f(i,j-1)) (a[i]!=b[j])

 

由于f(i,j)只和f(i-1,j-1), f(i-1,j)和f(i,j-1)有关, 而在计算f(i,j)时,

只要选择一个合适的顺序, 就可以保证这三项都已经计算出来了,

这样就可以计算出f(i,j). 这样一直推到f(len(a),len(b))就得到所要求的解了.

 1 #include <stdio.h>

 2 #include <string.h>

 3 int main() {

 4     int t,i,j,lena,lenb;

 5     char a[601],b[601];

 6     int map[601][601];

 7     while(EOF != scanf("%s %s",a,b)){

 8         lena=strlen(a);

 9         lenb=strlen(b);

10         memset(map , 0 ,sizeof(map));

11         for(i=1;i<=lena;i++){

12             for(j=1;j<=lenb;j++){

13                 if(a[i-1] == b[j-1])

14                     map[i][j] = map[i-1][j-1] + 1;

15                 else if(a[i-1] != b[j-1]){

16                     int temp = map[i-1][j];

17                     if(map[i][j-1] > temp)

18                         temp = map[i][j-1];

19                     map[i][j] = temp;

20                 }

21             }

22         }

23         printf("%d\n",map[lena][lenb]);

24     }

25     return 0;

26 }

HDOJ1003

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence.

For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 1 #include <stdio.h>

 2 #define MAX 100005

 3 int nNum[MAX];

 4 int start, end;

 5 //算法思路:用sum和max来记录临时最大和最终最大

 6 //而start end ts te 分别用来记录起点 终点 临时起始点 终点

 7 int cal(int len){

 8     int max, i, sum, ts, te;

 9     sum = max = -9999999; //让max和sum开始置于无限小

10     for(i = 0; i < len; i++){

11         if(sum < 0){

12             if(nNum[i] > sum){

13                 sum = nNum[i];

14                 ts = te = i;

15                 if(max < sum){

16                     max= sum;

17                     start = ts;

18                     end = te;

19                 }

20             }

21         }else{

22             sum += nNum[i];

23             te = i;

24             if(sum > max){

25                 max = sum;

26                 start = ts;

27                 end = te;

28             }

29         }

30     }

31     return max;

32 }

33 int main(){

34     int t, i, n, j;

35     while(EOF != scanf("%d", &t)){

36             for(i = 1; i <= t; i++){

37             scanf("%d", &n);

38             for(j = 0; j < n; j++)

39                 scanf("%d", &nNum[j]);

40             int max = cal(n);

41             printf("Case %d:\n", i);

42             printf("%d %d %d\n", max, start + 1, end + 1);

43             if(i != t)

44                 printf("\n");

45         }

46     }

47     return 0;

48 }