Zju1234 chopsticks-Dynamic Planning

Description:

One person invented a new method of using chopsticks. Each time three chopsticks were used, the chopsticks were of different lengths (A <= B <= C ), specify the "not moderate" of the three chopsticks as indicated by (A-B) ^ 2. Now there are n chopsticks with different lengths. I want to know the sum of the minimum and moderate values of M chopsticks. Input according to the chopsticks length in non-descending order.

Analysis:

ThisDynamic PlanningThe deformation is strange. The length of the three chopsticks, the longest one, is useless, but must exist.

Array a [I] indicates the length of the I-root chopsticks.Non-ascendingThe subscript starts from 1. UseStatus f [I] [J] indicates that j pairs are selected from the I-root chopsticks, and the minimum deviation is moderate..

It can be proved that if you choose the [I] root chopsticks as the shortest, then the short chopsticks must be a [I-1]. Therefore, note that bad (I) indicates that the I-root chopsticks are obtained as the shortest result.

Apparently, the initial status is:

F [I] [0] = 0; F [3] [1] = bad (3 ).

State transition equationF [I] [J] = min {f [I-1] [J], F [I-2] [J-1] + bad (I )}.

Note that f [I] [J] is invalid.

/*ZJU1234 Chopsticks*/#include 
 
  #include 
  
   #define N 5005#define K 1005#define INF 99999999#define clr(a) memset(a,0,sizeof(a))#define MIN(a,b) ((a)>(b)?(b):(a))#define POW(a) ((a)*(a))int n,m;int a[N];int f[N][K];int bad(int i){    return POW(a[i]-a[i-1]);}int main(){    int i,j,k,T;    int min;        scanf("%d",&T);    while(T--){        //init        clr(a);        clr(f);        min=INF;                //input        scanf("%d%d",&k,&n);        m=k+8;        for(i=n;i>=1;i--)            scanf("%d",&a[i]);                //DP        f[3][1]=bad(3);        for(i=4;i<=n;i++){            k=MIN(i/3,m);            for(j=1;j<=k;j++){                if(j*3==i) f[i][j]=INF;                else f[i][j]=f[i-1][j];                f[i][j]=MIN(f[i][j],f[i-2][j-1]+bad(i));                if(j==m) min=MIN(min,f[i][j]);            }        }                //output        printf("%d/n",min);            }        return 0;}