HDU1171 单调队列优化多重背包O(VN)_分组背包单调队列优化-优快云博客

本文链接：https://blog.youkuaiyun.com/ACM_DavidCN/article/details/5887406

本文介绍了一种解决多重背包问题的高效算法，并通过一个具体的竞赛题目进行了解析。该算法利用单调队列优化了传统的动态规划方法，实现了O(VN)的时间复杂度。

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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6218 Accepted Submission(s): 1996

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input

10 1

20 1

10 1

20 2

30 1

-1

Sample Output

20 10

40 40

显然是多重背包问题 , 这里使用单调队列优的 O(VN) 算法来解 .

代码如下 :

#include<stdio.h> #include<string.h> #define MAXN 105 struct Queue { int num,value; }que[250005]; int head,tail; int v[MAXN],w[MAXN],c[MAXN]; int dp[250005]; void enqueue (int x,int y) { while (head<=tail && que[tail].value<y) tail--; que[++tail].num=x;que[tail].value=y; } int main() { int i,j,d,sum,n,tempsum; while (scanf("%d",&n) && n>=0) { sum=0; for (i=1 ; i<=n ; ++i) { scanf("%d%d",&v[i],&c[i]); w[i]=v[i]; sum+=w[i]*c[i]; } tempsum=sum; sum/=2; for (i=1 ; i<=sum ; ++i) dp[i]=0; for (i=1 ; i<=n ; ++i) { if (c[i] > sum/w[i]) c[i]=sum/w[i]; for (d=0 ; d<w[i] ; ++d) { head=1;tail=0; for (j=0 ; j<=(sum-d)/w[i] ; ++j) { enqueue(j , dp[j*w[i]+d]-j*v[i]); while (que[head].num<j-c[i] && head<=tail) head++; dp[j*w[i]+d]=que[head].value+j*v[i]; } } } printf("%d %d/n",tempsum-dp[sum],dp[sum]); } return 0; }