【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.

 #include <stdio.h>
 #include <string.h>
 int main() {
     int t,i,j,lena,lenb;
     char a[],b[];
     int map[][];
     while(EOF != scanf("%s %s",a,b)){
         lena=strlen(a);
         lenb=strlen(b);
         memset(map ,  ,sizeof(map));
         for(i=;i<=lena;i++){
             for(j=;j<=lenb;j++){
                 if(a[i-] == b[j-])
                     map[i][j] = map[i-][j-] + ;
                 else if(a[i-] != b[j-]){
                     int temp = map[i-][j];
                     if(map[i][j-] > temp)
                         temp = map[i][j-];
                     map[i][j] = temp;
                 }
             }
         }
         printf("%d\n",map[lena][lenb]);
     }
     return ;
 }

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.

 #include <stdio.h>
 #define MAX 100005
 int nNum[MAX];
 int start, end;
 //算法思路:用sum和max来记录临时最大和最终最大
 //而start end ts te 分别用来记录起点 终点 临时起始点 终点
 int cal(int len){
     int max, i, sum, ts, te;
     sum = max = -; //让max和sum开始置于无限小
     for(i = ; i < len; i++){
         if(sum < ){
             if(nNum[i] > sum){
                 sum = nNum[i];
                 ts = te = i;
                 if(max < sum){
                     max= sum;
                     start = ts;
                     end = te;
                 }
             }
         }else{
             sum += nNum[i];
             te = i;
             if(sum > max){
                 max = sum;
                 start = ts;
                 end = te;
             }
         }
     }
     return max;
 }
 int main(){
     int t, i, n, j;
     while(EOF != scanf("%d", &t)){
             for(i = ; i <= t; i++){
             scanf("%d", &n);
             for(j = ; j < n; j++)
                 scanf("%d", &nNum[j]);
             int max = cal(n);
             printf("Case %d:\n", i);
             printf("%d %d %d\n", max, start + , end + );
             if(i != t)
                 printf("\n");
         }
     }
     return ;
 }