codility MinAbsSum

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For a given array A of N integers and a sequence S of N integers from the set {−1, 1}, we define val(A, S) as follows:

val(A, S) = |sum{ A[i]*S[i] for i = 0..N−1 }|

(Assume that the sum of zero elements equals zero.)

For a given array A, we are looking for such a sequence S that minimizes val(A,S).

Write a function:

int solution(int A[], int N);

that, given an array A of N integers, computes the minimum value of val(A,S) from all possible values of val(A,S) for all possible sequences S of N integers from the set {−1, 1}.

For example, given array:

A[0] = 1 A[1] = 5 A[2] = 2 A[3] = -2

your function should return 0, since for S = [−1, 1, −1, 1], val(A, S) = 0, which is the minimum possible value.

Assume that:

• N is an integer within the range [0..20,000];
• each element of array A is an integer within the range [−100..100].

Complexity:

• expected worst-case time complexity is O(N*max(abs(A))2);
• expected worst-case space complexity is O(N+sum(abs(A))), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

题目大意：给n个数，每个数的范围在-100到100之间，然后对于每个数，我们可以选着乘上-1或1，然后让你求出这些数的和的绝对值的最小值。

这题，可以转化成背包问题。

很明显，我们对这些数求和得sum ，用多重背包求得接近<=sum/2的最大值ans,那么sum-ans-ans就是结果了....这里不好语言描述，不过想一想还是很正确的.

把多重背包转换成01背包求。时间复杂度：O(V*Σlog n[i])  这里的V=sum/2

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int dp[1000010];
int V;
inline void dp1(int cost,int weight){
    for(int v=V;v>=cost;v--)
        dp[v]=max(dp[v],dp[v-cost]+weight);
}
inline void dp2(int cost,int weight){
    for(int v=cost;v<=V;v++){
        dp[v]=max(dp[v],dp[v-cost]+weight);
    }
}
inline void dp3(int cost,int weight,int num){
    if(cost*num>=V){
        dp2(cost,weight);
        return ;
    }
    int  k=1;
    while(k<num){
        dp1(k*cost,k*weight);
        num-=k;
        k*=2;
    }
    dp1(num*cost,num*weight);
}
int cnt[102];
int solution(vector<int> &A) {
    // write your code in C++11 (g++ 4.8.2)
    int n=A.size();
    int sum,m;
    sum=m=0;
    dp[0]=0;
    for(int i=0;i<n;i++){
        int x=(A[i]>=0)?A[i]:(-A[i]);
        sum+=x;
        cnt[x]++;
        if(x>m)m=x;
    }
    if(m==0)return 0;
    V=(sum>>1);
    for(int i=1;i<=m;i++){
        if(cnt[i])dp3(i,i,cnt[i]);
    }
    return (sum-(dp[V]<<1));
    
}

 more about 多重背包：http://love-oriented.com/pack/P03.html

codility MinAbsSum

标签：lock   color   ble   abs   [1]   debugging   with   sage   pre   

原文地址：http://www.cnblogs.com/pk28/p/6031474.html