一道好的DP

Consumer Submit: 786 Accepted:169 Time Limit: 2000MS Memory Limit: 65536K Description FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money. Input The first line will contain two integers, n (the number of boxes 1 <= n <= 50), w (the amount of money FJ has, 1 <= w <= 100000) Then n lines follow. Each line contains the following number pi (the price of the ith box 1<=pi<=1000), mi (1<=mi<=10 the number goods ith box can carry), and mi pairs of numbers, the price cj (1<=cj<=100), the value vj(1<=vj<=1000000) Output For each test case, output the maximum value FJ can get Sample Input 3 800 300 2 30 50 25 80 600 1 50 130 400 3 40 70 30 40 35 60 Sample Output 210 http://acm.scs.bupt.cn/onlinejudge/showproblem.php?problem_id=1848 #include #include #include #include using namespace std; int dp[100010],f[100010],V,N,m,n,cost,weight; int main() { while(scanf("%d%d",&N,&V)!=EOF) { memset(f,0,sizeof(f)); memset(dp,0,sizeof(dp)); for(int i=0;i=cost+m;v--)//对每一组/1背包 dp[v]=max(dp[v],dp[v-cost]+weight); } for(int k=cost;k<=V;k++)//如果更新了就赋给原dp if(dp[k]>f[k])f[k]=dp[k]; } printf("%d\n",f[V]); } return 0; } 还可以把有附件的背包转换为分组背包来做！（暂时TLE有待优化） #include #include #include #include using namespace std; int n,m,N,V,f[100010],dp[100010],total[100]; struct bag { int cost,weight; }New[60][100]; int main() { int cost,weight; while(scanf("%d%d",&N,&V)!=EOF) { memset(total,0,sizeof(total)); int mmin=1<<30; for(int i=0;icost)mmin=cost; for(int v=V;v>=m+cost;v--) dp[v]=max(dp[v],dp[v-cost]+weight); } for(int v=2;v<=V;v++) { if(dp[v]==dp[v-1])continue; New[i][total[i]].cost=v; New[i][total[i]++].weight=dp[v]; } } memset(f,0,sizeof(f)); for(int i=0;i=mmin;v--) { int mmax=0; for(int k=0;kmmax)mmax=f[v-New[i][k].cost]+New[i][k].weight; } if(f[v]