1570. Eating High

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;

}