Longest Common Subsequence && Substirng【公共子序列及子字符串】【经典dp】
2016-12-11 16:42
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题目1:Longest Common Subsequence
内容:给出a,b俩个字符串,求它们最长的公共子序列的长度!
思路:子序列:将字符串中任意位置的字符挑选出来组成的且顺序保持原顺序的字符串为子序列!
dp[i][j]表示一种状态:a字符串中到第i个位置和b字符串中到第j个位置此时的公共长度的最优解!
dp[][]一共4种:dp[i-1][j-1],dp[i-1][j],dp[i][j-1],dp[i][j];
所以当a[i] == b[i]时,只需dp[i][j] = dp[i-1][j-1]+1 (因为dp[i-1][j-1]为上一次相等时的最优解)
a[i] != b[i]时,只需dp[i][j] = max(dp[i-1][j],dp[i][j-1]) (从上一次不相等中的最优解挑选个最优解)
代码:
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn];
int main()
{
char a[maxn],b[maxn];
while(scanf("%s%s",a,b)!=EOF)
{
memset(dp,0,sizeof(dp));
int n = strlen(a),m = strlen(b);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(a[i-1] == b[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
printf("%d\n",dp
[m]);
}
return 0;
}
题目2:Longest Common Substirng
内容:a,b俩字符串,求公共连续的子字符串的长度!
思路:同上,
当a[i] == b[i]时,dp[i][j] = dp[i-1][j-1]+1
但是当a[i] != b[i]时,就不同了,因为要连续的字符串,所以直接将当前dp[i][j]切断即可,即dp[i][j] = 0;
最后从dp[][]中筛选出最大值即为最长公共连续子字符串了!
代码:
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn];
int main()
{
char a[maxn],b[maxn];
while(scanf("%s%s",a,b)!=EOF)
{
memset(dp,0,sizeof(dp));
int n = strlen(a),m = strlen(b),ans = 0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(a[i-1] == b[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = 0;
ans = max(ans,dp[i][j]);
}
printf("%d\n",ans);
}
return 0;
}
内容:给出a,b俩个字符串,求它们最长的公共子序列的长度!
思路:子序列:将字符串中任意位置的字符挑选出来组成的且顺序保持原顺序的字符串为子序列!
dp[i][j]表示一种状态:a字符串中到第i个位置和b字符串中到第j个位置此时的公共长度的最优解!
dp[][]一共4种:dp[i-1][j-1],dp[i-1][j],dp[i][j-1],dp[i][j];
所以当a[i] == b[i]时,只需dp[i][j] = dp[i-1][j-1]+1 (因为dp[i-1][j-1]为上一次相等时的最优解)
a[i] != b[i]时,只需dp[i][j] = max(dp[i-1][j],dp[i][j-1]) (从上一次不相等中的最优解挑选个最优解)
代码:
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn];
int main()
{
char a[maxn],b[maxn];
while(scanf("%s%s",a,b)!=EOF)
{
memset(dp,0,sizeof(dp));
int n = strlen(a),m = strlen(b);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(a[i-1] == b[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
printf("%d\n",dp
[m]);
}
return 0;
}
题目2:Longest Common Substirng
内容:a,b俩字符串,求公共连续的子字符串的长度!
思路:同上,
当a[i] == b[i]时,dp[i][j] = dp[i-1][j-1]+1
但是当a[i] != b[i]时,就不同了,因为要连续的字符串,所以直接将当前dp[i][j]切断即可,即dp[i][j] = 0;
最后从dp[][]中筛选出最大值即为最长公共连续子字符串了!
代码:
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn];
int main()
{
char a[maxn],b[maxn];
while(scanf("%s%s",a,b)!=EOF)
{
memset(dp,0,sizeof(dp));
int n = strlen(a),m = strlen(b),ans = 0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(a[i-1] == b[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = 0;
ans = max(ans,dp[i][j]);
}
printf("%d\n",ans);
}
return 0;
}
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