hdu 1423 Greatest Common Increasing Subsequence (最长上升子序列)
2013-08-06 14:16
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1、http://acm.hdu.edu.cn/showproblem.php?pid=1423
参考百度文库http://wenku.baidu.com/view/3e78f223aaea998fcc220ea0.html
2、题目大意:
求两个字符串的最大上升子序列,LCIS解决即可
Total Submission(s): 2116 Accepted Submission(s): 638
[align=left]Problem Description[/align]
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.
[align=left]Input[/align]
Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.
[align=left]Output[/align]
output print L - the length of the greatest common increasing subsequence of both sequences.
[align=left]Sample Input[/align]
[align=left]Sample Output[/align]
3、代码:
参考百度文库http://wenku.baidu.com/view/3e78f223aaea998fcc220ea0.html
2、题目大意:
求两个字符串的最大上升子序列,LCIS解决即可
Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2116 Accepted Submission(s): 638
[align=left]Problem Description[/align]
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.
[align=left]Input[/align]
Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.
[align=left]Output[/align]
output print L - the length of the greatest common increasing subsequence of both sequences.
[align=left]Sample Input[/align]
1 5 1 4 2 5 -12 4 -12 1 2 4
[align=left]Sample Output[/align]
2
3、代码:
#include<stdio.h> #include<algorithm> #include<cstring> using namespace std; int main() { int t,m,n,a[600],b[600],dp[601][601],i,j,max; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=0;i<n;i++) scanf("%d",&a[i]); scanf("%d",&m); for(i=0;i<m;i++) scanf("%d",&b[i]); memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) { max=0; for(j=0;j<m;j++) { dp[i+1][j+1]=dp[i][j+1]; if(a[i]>b[j]&&max<dp[i+1][j+1]) max=dp[i+1][j+1]; if(a[i]==b[j]) dp[i+1][j+1]=max+1; } } //printf("%d\n",m); //for(i=0;i<=m;i++) //printf("%d\n",dp [i]); max=*max_element(dp ,dp +m+1); printf("%d\n",max); if(t!=0) printf("\n"); } return 0; }
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