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NOIP1999 拦截导弹

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NOIP1999 拦截导弹 

题目描述 Description

    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 能高于前一发的高度。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。

  

输入描述 Input Description

输入导弹依次飞来的高度(雷达给出的高度数据是不大于30000的正整数)

  

输出描述 Output Description

输出这套系统最多能拦截多少导弹,如果要拦截所有导弹最少要配备多少套这种导弹拦截系统。

样例输入 Sample Input

389 207 155 300 299 170 158 65 

样例输出 Sample Output

6

2

数据范围及提示 Data Size & Hint

导弹的高度<=30000,导弹个数<=20

 

 

 

 

思路

这道题的突破口在于问题的转化= =

先是第一问: “最多能拦截的导弹数”。根据题目中导弹的毛病,可以把问题转化成经典的最长不上升子序列,动态规划O(cnt2)完美解决;

再看第二问: “最少需要的系统数”。顺延上一问的思路,“不可能由之前已经确定的系统拦截下来的导弹”应当添加到当前位置前的最长上升子序列中。同样是经典问题,动态规划O(cnt2)

最后稍微得瑟一下→_→在没有优化的情况下,这段代码在发帖前包揽了wikioi上“最快”“最短”以及“内存最少”三项指标~(好吧是因为这题太水了)

 

 1 #include <cstdio>
 2 using namespace std;
 3 int H[22], Suc[22]={1}, Cnt[22]={1}, mS=0, mC=0;
 4 int main()
 5 {
 6     freopen("testin","r",stdin);
 7     freopen("testout","w",stdout);
 8     int i,j,t,cnt=0;
 9     while(scanf("%d", &t)==1)
10         H[cnt++]=t;
11     for(i=1;i<cnt;++i) {
12         Suc[i]=Cnt[i]=1;
13         for(j=i-1;j>=0;--j) {
14             if(H[j]>=H[i]&&Suc[j]>=Suc[i])
15                 Suc[i]=Suc[j]+1;
16             if(H[j]<H[i]&&Cnt[j]>=Cnt[i])
17                 Cnt[i]=Cnt[j]+1;
18         }
19         if(Suc[i]>mS)mS=Suc[i];
20         if(Cnt[i]>mC)mC=Cnt[i];
21     }
22     printf("%d\n%d", mS, mC);
23     return 0;
24 }