zoj 3812 We Need Medicine (dp 状压)

先贴一下转载的思路和代码，，，：http://blog.csdn.net/qian99/article/details/39138329

状压dp博大精深啊，以后看到n<=50都可以往状压上想，orz

We Need Medicine

Time Limit: 10 Seconds Memory Limit: 65536 KB Special Judge

A terrible disease broke out! The disease was caused by a new type of virus, which will lead to lethal lymphoedema symptom. For convenience, it was named LL virus.

After several weeks of research, the scientists found the LL virus highly lethal and infectious. But more importantly, it has a long incubation period. Many victims were unaware of being infected until everything was too late. To prevent from the apocalypse, we need medicine!

Fortunately, after another several weeks of research, the scientists have finished the analysis of the LL virus. You need write a program to help them to produce the medicine.

The scientists provide you N kinds of chemical substances. For each substance, you can either use it exact W_i milligrams in a medicine, or not use it. Each selected substance will add T_i points of therapeutic effect value (TEV) to the medicine.

The LL virus has Q different variants. For each variant, you need design a medicine whose total weight equals to M_i milligrams and total TEV equals to S_i points. Since the LL virus is spreading rapidly, you should start to solve this problem as soon as possible!

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N (1 <= N <= 400) and Q (1 <= Q <= 400).

For the next N lines, each line contains two integers W_i (1 <= W_i <= 50) and T_i (1 <= T_i <= 200000).

Then followed by Q lines, each line contains two integers M_i (1 <= M_i <= 50) and S_i (1 <= S_i <= 200000).

Output

For each test case, output Q lines. For the i-th line, output the indexes (1-based) of chemical substances in the i-th medicine, separated by a space. If there are multiple solutions, output any one. If there is no solution, output "No solution!" instead.

Sample Input

Sample Output

1 3

3 2 1

No solution!

Author: JIANG, Kai Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

转自：http://blog.csdn.net/qian99/article/details/39138329

题意：给出n个物品，每个物品有两种属性Wi,Ti，有q组查询，每组查询要求在n个物品中选出一些，并使得两个属性的和为Mi，Si。

思路：刚开始看感觉是神题，后来仔细想了想，其实本质上就是个背包。最裸着写的话，那么就是dp[i][j][k]表示使用前i个物品，是否可以凑出第一个属性j，第二个属性k，要输出方案的话记录一下路径就可以了。一开始这么写了一发，加了一些乱七八糟的优化，还是会T。虽然这题时限还算宽，但这么写复杂度还是太高了。考虑到第一个属性最多只有50，那么可以用一个二进制数来表示是否能凑出第一个属性的情况，即：第i位为1表示可以凑出i。使用这种方法的好处是对于物品i可以直接算出第一种属性的组合情况，枚举一下新增的位，更新一下结果就行了。

 #include<iostream>

 #include<cstring>

 #include<cstdlib>

 #include<cstdio>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<map>

 #define N 410

 #define M 200010

 #define mod 6

 #define mod2 100000000

 #define ll long long

 #define ull unsigned long long

 #define maxi(a,b) (a)>(b)? (a) : (b)

 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int T;

 int n,q;

 int w[N],t[N];

 int m,s;

 ull f[M];

 int ans[M][];

 map<ull,int>mt;

 void ini1()

 {

     int i;

     for(i=;i<=;i++){

         mt[ (1ll<<(i-1ll)) ]=i;

     }

 }

 void ini()

 {

     int i,j;

     ull k,x;

     ull v;

     scanf("%d%d",&n,&q);

     memset(ans,,sizeof(ans));

     memset(f,,sizeof(f));

     for(i=;i<=n;i++){

         scanf("%d%d",&w[i],&t[i]);

     }

     f[]=;

     for(i=;i<=n;i++){

         for(j=;j>=t[i];j--){

             v=f[j];

             f[j] |= (f[j-t[i]]<<w[i]) & ( (1ll<<)- );

             for(k=v ^ f[j];k>;k &= (k-) )

             {

                 x=(k ^(k-))&k;

                 ans[j][ mt[x]- ]=i;

             }

         }

     }

 }

 void solve()

 {

     int te;

     while(q--)

     {

         scanf("%d%d",&m,&s);

         if(ans[s][m]==){

             printf("No solution!\n");

             continue;

         }

         else{

             printf("%d",ans[s][m]);

             te=ans[s][m];

             m-=w[te];

             s-=t[te];

             while(m!=)

             {

                 printf(" %d",ans[s][m]);

                 te=ans[s][m];

                 m-=w[te];

                 s-=t[te];

             }

             printf("\n");

         }

     }

 }

 void out()

 {

     //printf("%lld\n",ans);

     //cout<<ans<<endl;

 }

 int main()

 {

     ini1();

    // freopen("data.in","r",stdin);

     scanf("%d",&T);

     for(int cnt=;cnt<=T;cnt++)

    // while(T--)

     //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)

     {

         ini();

         solve();

         //out();

     }

     return ;

 }

再贴一份自己加过注释的，状压dp博大精深啊：

 #include<iostream>

 #include<cstring>

 #include<cstdlib>

 #include<cstdio>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<map>

 #define N 410

 #define M 200010

 #define mod 6

 #define mod2 100000000

 #define ll long long

 #define ull unsigned long long

 #define maxi(a,b) (a)>(b)? (a) : (b)

 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int T;

 int n,q;

 int w[N],t[N];

 int m,s;

 ull f[M];

 int ans[M][];

 map<ull,int>mt;

 void ini1()

 {

     int i;

     for(i=;i<=;i++){

         mt[ (1ll<<(i-1ll)) ]=i;

         //printf(" i=%d mt=%I64d\n",i,(1ll<<(i-1ll)));

     }

 }

 void ini()

 {

     int i,j;

     ull k,x;

     ull v,te;

     scanf("%d%d",&n,&q);

     memset(ans,,sizeof(ans));

     memset(f,,sizeof(f));

     for(i=;i<=n;i++){

         scanf("%d%d",&w[i],&t[i]);

     }

     f[]=;

     for(i=;i<=n;i++){

         //printf(" i=%d\n",i);

         //for(j=200000;j>=t[i];j--){

         for(j=;j>=t[i];j--){

             v=f[j];             //j原来存的值

             te=(f[j-t[i]]<<w[i]);

             f[j] |= (f[j-t[i]]<<w[i]) ;     //j原来存的值+转移后存的值

           //  printf("  j=%d v=%I64d f=%I64d\n",j,v,f[j]);

             for(k=v ^ f[j];k>;k &= (k-) )  //k初始化为新增出来的值，然后不断将k最右边的1减掉

            //  for(k=te;k>0;k &= (k-1) )

             {

                 x=(k ^(k-))&k;         //取k最右边的1

                 ans[j][ mt[x]- ]=i;

                // printf("    k=%I64d x=%I64d mtx-1=%d ans=%d\n",k,x,mt[x]-1,ans[j][ mt[x]-1 ]);

             }

         }

     }

 }

 void solve()

 {

     int te;

     while(q--)

     {

         scanf("%d%d",&m,&s);

         if(ans[s][m]==){

             printf("No solution!\n");

             continue;

         }

         else{

             printf("%d",ans[s][m]);

             te=ans[s][m];

             m-=w[te];

             s-=t[te];

             while(m!=)

             {

                 printf(" %d",ans[s][m]);

                 te=ans[s][m];

                 m-=w[te];

                 s-=t[te];

             }

             printf("\n");

         }

     }

 }

 void out()

 {

     //printf("%lld\n",ans);

     //cout<<ans<<endl;

 }

 int main()

 {

     ini1();

    // freopen("data.in","r",stdin);

     //freopen("data.out","w",stdout);

     scanf("%d",&T);

     for(int cnt=;cnt<=T;cnt++)

    // while(T--)

     //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)

     {

         ini();

         solve();

         //out();

     }

     return ;

 }