【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[601],b[601];
    int map[601][601];
    while(EOF != scanf("%s %s",a,b)){
        lena=strlen(a);
        lenb=strlen(b);
        memset(map , 0 ,sizeof(map));
        for(i=1;i<=lena;i++){
            for(j=1;j<=lenb;j++){
                if(a[i-1] == b[j-1])
                    map[i][j] = map[i-1][j-1] + 1;
                else if(a[i-1] != b[j-1]){
                    int temp = map[i-1][j];
                    if(map[i][j-1] > temp)
                        temp = map[i][j-1];
                    map[i][j] = temp;
                }
            }
        }
        printf("%d
",map[lena][lenb]);
    }
    return 0;
}

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 = -9999999; //让max和sum开始置于无限小
    for(i = 0; i < len; i++){
        if(sum < 0){
            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 = 1; i <= t; i++){
            scanf("%d", &n);
            for(j = 0; j < n; j++)
                scanf("%d", &nNum[j]);
            int max = cal(n);
            printf("Case %d:
", i);
            printf("%d %d %d
", max, start + 1, end + 1);
            if(i != t)
                printf("
");
        }
    }
    return 0;
}