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-;
 const int MS=;

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

 LL w[MS],v[MS];

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

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

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-;
 const int MS=;

 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[<<(MS/)];

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

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

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