Mr. Panda and Crystal HDU

应该注意到假如物品的费用可以被缩小的话
那一定是>=材料中最小的物品的费用

所以可以先对每个物品求最小费用
然后再做一个完全背包

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define pll pair<ll,ll>#define mp(aa,bb) make_pair(aa,bb)#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  200+10;const int maxm = 10000+10;//Pretests passedint in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}int dis[maxn];int money[maxn];bool vis[maxn];struct node{    int x;    vector<pii>g;}no[maxn];vector<int>v[maxn];int dp[maxm];int n,m,k;int cal(int q){    int res = 0;    rsz(i,no[q].g)    {        pii p = no[q].g[i];        int a = p.X , b = p.Y;        res += dis[a]*b;    }    return res;}void dij(){    mst(vis,0);    priority_queue<pii>q;    r1(i,n)if(dis[i]<=m)q.push(mp(-dis[i],i));    while(!q.empty())    {        pii p = q.top();        q.pop();        int u = p.Y;        if(vis[u])continue;        vis[u] = 1;        rsz(i,v[u])        {            int idx = v[u][i];            int dist = cal(idx);            int nt = no[idx].x;            if(dis[nt] > dist)            {                dis[nt] = dist;                q.push(mp(-dist,nt));            }        }    }}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int t;    sd(t);    r1(cas,t)    {        printf("Case #%d: ",cas);//        int n,m,k;        sddd(m,n,k);        r1(i,n)        {            int op;            sd(op);            if(op)            {                int a,b;                sdd(a,b);                money[i] = b;                dis[i] = a;            }            else            {                int x;                sd(x);                money[i] = x;                dis[i] = m+1;            }            v[i].clear();        }        r1(i,k)        {            no[i].g.clear();            int x,y;            sdd(x,y);            no[i].x = x;            while(y--)            {                int a,b;                sdd(a,b);                no[i].g.pb(mp(a,b));                v[a].pb(i);            }        }        dij();        mst(dp,0);        for(int i=1;i<=n;++i)        {            for(int j = dis[i];j<=m;++j)            {                dp[j] = max(dp[j],dp[j - dis[i] ] + money[i]);            }        }        int ans = dp[m];        ansn();    }    return 0;}