Longest Ordered Subsequence
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
Sample Input
7
1 7 3 5 9 4 8
Sample Output
4 LIS最长上升子序列问题。（可加二分优化O(nlogn)）f[i]记录以当前值作为最后一个数时的最大长度。每枚举一个数都找前面比他小且f[i]最大的状态+1。f[i]=max(f[j])+1
#include<stdio.h>
#include<string.h> int f[],a[]; int max(int x,int y)
{
return x>y?x:y;
} int main()
{
int n,i,j;
scanf("%d",&n);
memset(a,,sizeof(a));
memset(f,,sizeof(f));
for(i=;i<=n;i++){
scanf("%d",&a[i]);
}
f[]=;
for(i=;i<=n;i++){
for(j=;j<i;j++){
if(a[j]<a[i]){
f[i]=max(f[i],f[j]);
}
}
f[i]++;
}
int ans=;
for(i=;i<=n;i++){
ans=max(ans,f[i]);
}
printf("%d\n",ans);
return ;
}
最少拦截系统
某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不能超过前一发的高度.某天,雷达捕捉到敌国的导弹来袭.由于该系统还在试用阶段,所以只有一套系统,因此有可能不能拦截所有的导弹.
怎么办呢?多搞几套系统呗!你说说倒蛮容易,成本呢?成本是个大问题啊.所以俺就到这里来求救了,请帮助计算一下最少需要多少套拦截系统.
怎么办呢?多搞几套系统呗!你说说倒蛮容易,成本呢?成本是个大问题啊.所以俺就到这里来求救了,请帮助计算一下最少需要多少套拦截系统.
Input
输入若干组数据.每组数据包括:导弹总个数(正整数),导弹依此飞来的高度(雷达给出的高度数据是不大于30000的正整数,用空格分隔)
Output
Output
对应每组数据输出拦截所有导弹最少要配备多少套这种导弹拦截系统.
Sample Input
8 389 207 155 300 299 170 158 65
Sample Output
2
贪心解决最少序列个数问题。数组记录每个序列的最小值，出现比他大的数存入新下标，数组长度即为个数。
#include<stdio.h>
#include<string.h> int a[]; int main()
{
int n,x,c,i,j;
while(~scanf("%d",&n)){
c=;
memset(a,,sizeof(a));
a[]=;
for(i=;i<=n;i++){
scanf("%d",&x);
for(j=;j<=c;j++){
if(a[j]>=x){
a[j]=x;
break;
}
if(j==c){
a[++c]=x;
break;
}
}
}
printf("%d\n",c);
}
return ;
}