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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 }
NOIP1999 拦截导弹