动态规划的一个入门问题：导弹防御

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动态规划入门，导弹防御问题。

实际上就是求一个最长的非递增序列。我的解答如下：

 #define _CRT_SECURE_NO_DEPRECATE/*keep away the warning for scanf_s(...) etc*/ #include <stdio.h> #define MAX 4000 int main() { int n; int missile[MAX],d[MAX]; /*missile[] indicates the height of each missile, *and d[] represets when the ith missile was intercepted,the max number for which the system can head off(including the ith missile)*/ int i,j; int dmax,first; /*keep a record of the max number of missiles can be intercepted,and the seq of the first missile to be intercepted*/ freopen("in.txt","r",stdin); while(scanf("%d",&n)==1) { dmax=0; first=0; for(i=0;i<n;i++) { scanf("%d",&missile[i]);/*the height value*/ d[i]=1;/*initial to 1 for each*/ } for(i=n-2;i>=0;i--)/*the loop from back to start*/ { for(j=i+1;j<n;j++) { if((missile[i]>missile[j]) && (d[i]<d[j]+1))/*decision condition*/ d[i]=d[j]+1; } } for(i=0;i<n;i++)/*find out the maximum number*/ { if(d[i]>dmax) { dmax=d[i]; first=i; //the first missile to be intercepted } } printf("%d/n%d",d[first],missile[first]);/*the output*/ for(i=first+1;i<n;i++) { if((missile[i]<=missile[first]) && (d[first]==(d[i]+1)))/*like a link list*/ { printf(" %d",missile[i]); first=i; } } printf("/n"); } return 0; }

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Input：

5
1000 23 52 89 10
1
200
7
8723 9999 1237 23 90 13 66
8
2000 1500 20 1200 1000 890 398 10

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Output：

3
1000 23 10
1
200
4
8723 1237 23 13
7
2000 1500 1200 1000 890 398 10