poj1080 Human Gene Functions (lcs变形)
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2022-07-12 08:58:56
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题意:给定两个字符串 str1 和 str2 ,在两个串中都可以插入空格,使两个串的长度最后相等,然后开始匹配,怎样插入空格由匹配规则得到的值最大。
分析:LCS的变形,dp[i][j]表示str1[1...i]和str2[1...j]两个子串的最大匹配值。与lcs类似可以分3个方向对其进行状态转移:
①由dp[i-1][j-1]转移过来,意味着str1[i]和str2[j]进行匹配,新增代价是str1[i]和str2[j]的匹配值。
②由dp[i-1][j]转移过来,意味着str1[1...i-1]和str2[1...j]已经进行过匹配,str1[i]只能和空字符匹配,新增代价是str1[i]和空字符的匹配值。
③由dp[i][j-1]转移过来,与②类似。
得到状态转移方程:
dp[i][j] = max(dp[i - 1][j - 1] + match[str1[i]][str2[j]],
max(dp[i - 1][j] + match[str1[i]][' '], dp[i][j - 1] + match[str2[j]][' ']));
代码如下:
#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
int match[256][256];
int dp[110][110];
char str1[110], str2[110];
void init()
{
match['A']['A'] = match['C']['C'] = match['G']['G'] = match['T']['T'] = 5;
match['A']['C'] = match['C']['A'] = match['A']['T'] = match['T']['A'] = match['T'][' '] = match[' ']['T'] = -1;
match['A']['G'] = match['G']['A'] = match['C']['T'] = match['T']['C'] = match['T']['G'] = match['G']['T'] = match[' ']['G'] = match['G'][' '] = -2;
match['C']['G'] = match['G']['C'] = match[' ']['A'] = match['A'][' '] = -3;
match['C'][' '] = match[' ']['C'] = -4;
}
int main()
{
//fstream cin("test.txt");
int k;
cin >> k;
init();
while (k--)
{
memset(dp, 0, sizeof(dp));
int len1, len2;
cin >> len1 >> str1+1 >> len2 >> str2+1;
dp[0][0] = 0;
for (int i = 1; i <= len2; i++)
dp[0][i] = dp[0][i - 1] + match[str2[i]][' ']; //注意这里一定要加上前一项
for (int i = 1; i <= len1; i++) //因为计算的是前i个字符和空字符的匹配值
dp[i][0] = dp[i - 1][0] + match[str1[i]][' '];
for (int i = 1; i <= len1; i++)
for (int j = 1; j <= len2; j++)
{
//dp[i][j] = dp[i - 1][j - 1] + match[str2[i - 1]][str1[j - 1]];
dp[i][j] = max(dp[i - 1][j - 1] + match[str1[i]][str2[j]],
max(dp[i - 1][j] + match[str1[i]][' '], dp[i][j - 1] + match[str2[j]][' ']));
}
cout << dp[len1][len2] << endl;
}
//system("pause");
return 0;
}
需要注意的是我在做这题的时候WA了一次,看了别人的代码才发现是初始化的时候写错了,我想当然地把dp[i][0]和dp[0][j]的值初始化为str1[i]、str2[j]与空字符的匹配值,实际应该是str1[1...i]、str2[1...j]与空字符的匹配值。只能说自己对dp的理解还是不够到位吧。
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