BZOJ-1010 Toy packing toy (slope optimization)

The main topic: The n number is divided into groups, and the number of each group in the original array should be continuous, each group will incur the cost of sum (i)-sum (j) +i-j-1-m,m is known constant. Minimum cost.

Topic Analysis: The definition of DP (i) represents the minimum cost of grouping the first I elements, and the state transition equation is obvious:

DP (i) =min (DP (j) +[sum (i)-sum (j) +i-j-1-m)]^2). Another f (i) =sum (i) +i, and the other G (i) =f (i) -1-m, the DP (i) can be organized into DP (i) =min (DP (j) +sum (j) ^2-2*g (i) *sum (j)) +g (i). It is clear that slope optimization is required.

The code is as follows:

# include<iostream># include<cstdio># include<cstring># include<algorithm>using namespace std;# define LL long longconst int n=50005;int n,m;int q[n]; LL A[n]; LL Dp[n];    ll Sum[n];void Read (ll &x) {char ch= "; while (ch< ' 0 ' | |    Ch> ' 9 ') Ch=getchar ();    x=0;        while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 ';    Ch=getchar ();    }}void init () {sum[0]=0;        for (int i=1;i<=n;++i) {read (a[i]);    SUM[I]=A[I]+SUM[I-1]; } for (int i=1;i<=n;++i) sum[i]+=i;} LL Getson (int k,int j) {return dp[j]-dp[k]+ (Sum[j]+sum[k]) * (Sum[j]-sum[k]);} LL getmother (int k,int j) {return (sum[j]-sum[k]);} Double getk (int i,int j) {return (double) Getson (I,J)/(double) getmother (I,J);} LL TODP (int j,int i) {return dp[j]+ (sum[i]-sum[j]-m-1) * (sum[i]-sum[j]-m-1);}    LL solve () {int head=0,tail=-1;    q[++tail]=0;    dp[0]=0; for (int i=1;i<=n;++i) {while (HEAD+1&LT;=TAIL&AMP;&AMP;GETK (q[head],q[head+1]) <=sum[i]-m-1) ++head;        DP[I]=TODP (Q[head],i);        while (HEAD+1&LT;=TAIL&AMP;&AMP;GETK (Q[tail-1],q[tail]) &GT;GETK (q[tail],i))--tail;    Q[++tail]=i; } return dp[n];}        int main () {while (~scanf ("%d%d", &n,&m)) {init ();    printf ("%lld\n", Solve ()); } return 0;}

　　

BZOJ-1010 Toy packing toy (slope optimization)