动态规划（2）
文章目录
1、聪明的寻宝人
c
复制代码
#include <iostream>
using namespace std;
void MaxValue(int values[], int weights[], int n, int m) {
int dp[21][51] = {0};
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (weights[i - 1] <= j) {
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weights[i - 1]] + values[i - 1]);
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
cout << dp[n][m] << endl;
}
2、基因检测
c
复制代码
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
void Similar(char *str1, char *str2) {
int m = strlen(str1);
int n = strlen(str2);
int dp[51][51] = {0};
int maxLen = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (str1[i - 1] == str2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
maxLen = max(maxLen, dp[i][j]);
} else {
dp[i][j] = 0;
}
}
}
cout << maxLen ;
}
3、药剂稀释
c
复制代码
#include <algorithm>
using namespace std;
void Cal(double arr[],int n)
{
/********** Begin **********/
//补充代码完成任务
int dp[n];
for(int i=0;i<n;i++) dp[i]=1;
for(int i=n-2;i>=0;i--){
for(int j=i+1;j<n;j++){
if(arr[i]>=arr[j]) dp[i]=dp[i]<(dp[j]+1)?(dp[j]+1):dp[i];
}
}
int max=1;
for(int i=0;i<n;i++){
if(max<dp[i]) max=dp[i];
}
printf("%d",max);
/********** End **********/
}
4、找相似串
c
复制代码
#include <iostream>
#include <cstring>
using namespace std;
const int MAX=60;
void Similar()
{
/********** Begin **********/
char s[MAX];
int n,end;
cin >> s>>n;//读取主串和子串个数
int len_s = strlen(s);
char arr[20][MAX];
int caozuo[20];//存操作次数
int dp[MAX][MAX];//用数组dp[i][j]表示，子串从1-i转换到主串的操作数。
for (int i = 0; i < n; i++)//读取子串
{
cin>>arr[i];
}
for (int i = 0; i < len_s; i++)
{
dp[0][i] = i; //处理边界
}
for (int k = 0; k < n; k++)//第k个子串
{
int len = strlen(arr[k]);//子串长度
//初始化
for (int j = 0; j < len; j++)
dp[j][0] = j;
for (int i = 1; i < len_s; i++)//i为主串下标
{
for (int j = 1; j < len; j++)//j为子串下标
{
if (s[i] == arr[k][j])
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;
}
}
caozuo[k] = dp[len_s-1][len-1];//存每个子串的最小操作数
}
end = caozuo[0];
for (int i = 1; i < n; i++)
end = min(end, caozuo[i]); //找到最小操作数
for (int i = 0; i < n; i++)
{
if (caozuo[i] == end)
cout << arr[i] << endl; //输出对应串
}
/********** End **********/
}