1570. Eating High 夜

http://acm.timus.ru/problem.aspx?space=1&num=1570

此题如果只求最少花费的话，就可以简单的dp或是背包就可以了

难就难在在选择路径上有困难  应该在记录路径时 记下所有可能是最优的路径 排除一定不是最优的路径

然后选择一条最优的总路径

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include <iomanip>
using namespace std;
#define LL long long
const double eps=1e-6;
const int INF=0x3f3f3f3f;
const int N=40005;
const int M=105;
struct node
{
    string name;
    int cost;
    int weight;
}mem[M];
int dp[N],num[N],f[N];
vector<int>vtf[N],vtnum[N];
int sum[M];
int n,m;
bool cmp(node x,node y)
{
    return x.cost<y.cost;
}
void packComplete(int k,int cost,int weight,int V)
{
    for(int v=cost;v<=V;++v)
    {
        if(v-cost>=m)
        break;
        if(dp[v-cost]+weight>dp[v])
        continue;
        if(dp[v-cost]+weight<dp[v])
        {
            vtf[v].clear();vtnum[v].clear();
            dp[v]=dp[v-cost]+weight;
            num[v]=num[v-cost]+((f[v-cost]==k)?0:1);
            vtf[v].push_back(k);vtnum[v].push_back(num[v]);
            f[v]=k;
        }else if(num[v-cost]+((f[v-cost]==k)?0:1)>num[v])
        {
            num[v]=num[v-cost]+((f[v-cost]==k)?0:1);
            vtf[v].push_back(k);vtnum[v].push_back(num[v]);
            f[v]=k;
        }
    }
}
int main()
{
    //freopen("data.in","r",stdin);
    while(cin>>n>>m)
    {
        m=int(m*1000+eps);
        int V=N-5;
        for(int i=1;i<=n;++i)
        {
            double ftmp;
            cin>>mem[i].name>>mem[i].weight>>ftmp;
            mem[i].cost=int(ftmp*1000+eps);
        }
        sort(mem+1,mem+n+1,cmp);
        memset(f,0,sizeof(f));
        memset(num,0,sizeof(num));
        for(int i=1;i<=V;++i)
        dp[i]=INF;
        dp[0]=0;
        for(int i=1;i<=n;++i)
        packComplete(i,mem[i].cost,mem[i].weight,V);
        memset(sum,0,sizeof(sum));
        int k=V;
        for(int i=m;i<V;++i)
        if(dp[i]<dp[k]||(dp[i]==dp[k]&&num[i]>num[k]))
        k=i;
        cout<<dp[k]<<endl;
        int pnum=num[k],pk=-1;
        while(k!=0)
        {
            for(unsigned int i=0;i<vtf[k].size();++i)
            {
                if(vtnum[k][i]+((vtf[k][i]==pk||pk==-1)?0:1)==pnum)
                {
                    pnum=vtnum[k][i];
                    pk=vtf[k][i];
                    ++sum[pk];
                    k=k-mem[pk].cost;
                    break;
                }
            }
        }
        for(int i=1;i<=n;++i)
        if(sum[i]!=0)
        cout<<mem[i].name<<" "<<sum[i]<<endl;
    }
    return 0;
}