hdu1423Greatest Common Increasing Subsequence(最长公共递增子序列)
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Greatest Common Increasing Subsequence
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 10 Accepted Submission(s) : 3
151 4 2 5 -124-12 1 2 4
2
ACM暑期集训队练习赛(二)
题意:
求两个数组的最长公共递增子序列
#include<cstdio>
#include<cstring>
int a[10000],b[10000],dp[1000][1000];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int x1,x2,i,j,max1;
scanf("%d",&x1);
for(i=0;i<x1;i++)
scanf("%d",&a[i]);
scanf("%d",&x2);
for(i=0;i<x2;i++)
scanf("%d",&b[i]);
memset(dp,0,sizeof(dp));
for(i=1;i<=x1;i++)
{
max1=0;
for(j=1;j<=x2;j++)
{
dp[i][j]=dp[i-1][j];//至少等于前面的
if(a[i-1]>b[j-1]&&max1<dp[i-1][j])//有可能在公共子序列里面;
max1=dp[i-1][j];
if(a[i-1]==b[j-1])//相等就说明是递增子序列里面的。
dp[i][j]=max1+1;
}
}
max1=0;
for(i=0;i<=x1;i++)//找出较大的
for(j=0;j<=x2;j++)
if(max1<dp[i][j])
max1=dp[i][j];
printf("%d\n",max1);
if(n)printf("\n");//注意就是最后的没有换行。
}
}
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