CSU OJ PID=1514: Packs 超大背包问题，折半枚举+二分查找。

1514: Packs

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 61  Solved: 4
[Submit][Status][Web Board]

Description

Give you n packs, each of it has a value v and a weight w. Now you should find some packs, and the total of these value is max, total of these weight is equal to m.

Input

First line is a number T( T ≤ 5) represent the test cases.
Then for each set of cases, first line is n (1 ≤ n ≤ 40) and m (1 ≤ m < 2^31), follow n line each is Wi (1 ≤ Wi < 2^31) and Vi (-2^31 < Vi < 2^31).

Output

Each case a line for max value.(Each set of inputs to ensure the solvability)

Sample Input

2
3 3
1 1
2 2
3 4
5 2
1 -5
1 -8
1 0
1 -2
1 5

Sample Output

4
5

注意：这里要求是满足总重量必须固定，而不是接近总重量。这就成了直接折半枚举，map大法就行了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long LL;
const long long INF=99999999999999999LL;
const int EXP=1e-5;
const int MS=42;

map<LL,LL> mp;
int n;
LL W;

LL w[MS],v[MS];

void solve()
{
    mp.clear();
    int n1=n/2;
    for(int i=0;i<(1<<n1);i++)
    {
        LL sw=0,sv=0;
        for(int j=0;j<n1;j++)
        {
            if((i>>j)&1)
            {
                sw+=w[j];
                sv+=v[j];
            }
        }
        mp[sw]=max(mp[sw],sv);
    }
    LL ans=-INF;
    for(int i=0;i< 1<<(n-n1);i++)
    {
        LL sw=0,sv=0;
        for(int j=0;j<(n-n1);j++)
        {
            if((i>>j)&1)
            {
                sw+=w[n1+j];
                sv+=v[n1+j];
            }
        }
       if(mp.count(W-sw))
            ans=max(ans,mp[W-sw]+sv);
    }
    printf("%lld
",ans);
}


int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%lld",&n,&W);
        for(int i=0;i<n;i++)
            scanf("%lld%lld",&w[i],&v[i]);
        solve();
    }
    return 0;
}

如果是重量不大于 W ，那么就要折半枚举+二分查找。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long LL;
const int INF=0x4fffffff;
const int EXP=1e-5;
const int MS=40;

int n;
LL W;
LL w[MS],v[MS];
struct node
{
    LL w,v;
    bool operator <(const node &a) const   //注意一定要加上const
    {
        return w<a.w||(w==a.w&&v<a.v);
    }
}nodes[1<<(MS/2)];

LL find(LL w,int cnt)
{
    int l=0,r=cnt;
    while(r-l>1)         //  左闭右开区间处理起来更方便。
    {
        int mid=(l+r)/2;
        if(nodes[mid].w<=w)
            l=mid;
        else
            r=mid;
    }
    return nodes[l].v;
}

void solve()
{
    int n1=n/2;
    int cnt=0;
    for(int i=0;i<(1<<n1);i++)
    {
        LL sw=0,sv=0;
        for(int j=0;j<n1;j++)
        {
            if((i>>j)&1)
            {
                sw+=w[j];
                sv+=v[j];
            }
        }
        nodes[cnt].w=sw;
        nodes[cnt++].v=sv;
    }
    sort(nodes,nodes+cnt);
    int last=0;
    for(int i=0;i<cnt;i++)
    {
        if(nodes[last].v<nodes[i].v)
        {
            nodes[++last]=nodes[i];
        }
    }
    cnt=last+1;
    LL ans=0;
    for(int i=0;i< 1<<(n-n1);i++)
    {
        LL sw=0,sv=0;
        for(int j=0;j<(n-n1);j++)
        {
            if((i>>j)&1)
            {
                sw+=w[n1+j];
                sv+=v[n1+j];
            }
        }
        if(sw<=W)
        {
            LL tv=find(W-sw,cnt);
            ans=max(ans,sv+tv);
        }
    }
    printf("%lld
",ans);
}


int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%lld",&n,&W);
        for(int i=0;i<n;i++)
            scanf("%lld%lld",&w[i],&v[i]);
        solve();
    }
    return 0;
}