昂贵的聘礼 POJ

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的确是道好题考验建图能力

把可交换的价格看作边权如果a物品可以用b物品加c金币换到就建一条从a到b且权值为c的边

其实不管怎么买卖手中的东西始终只有一个所以从各个物品到水晶球与从水晶球到各个物品其实含义一样反向以水晶球为起点就好

跑完最短路后 dis[i]就代表开始先买i物品一路买卖交易直到换取酋长水晶球然后用所有dis[i]+val[i](物品本身价值) 与酋长水晶球的价值比较即可

#include <stdio.h>#include <queue>#include <cstring>#include <algorithm>using namespace std;struct nodeI{    int p;    int l;};struct nodeII{    int v;    int w;    int next;};struct nodeIII{    friend bool operator < (nodeIII n1,nodeIII n2)    {        return n1.w>n2.w;    }    int v;    int w;};nodeI point[110];nodeII edge[10010];priority_queue <nodeIII> que;int first[110],dis[110],book[110];int lim,n,num,ans;void addedge(int u,int v,int w){    edge[num].v=v;    edge[num].w=w;    edge[num].next=first[u];    first[u]=num++;    return;}void calculate(int ll,int rr){    nodeIII cur,tem;    int i,u,v,w;    while(!que.empty()) que.pop();    memset(dis,0x3f,sizeof(dis));    memset(book,0,sizeof(book));    tem.v=1,tem.w=0;    que.push(tem);    dis[1]=0;    while(!que.empty())    {        cur=que.top();        que.pop();        u=cur.v;        if(book[u]==1) continue;        book[u]=1;        for(i=first[u];i!=-1;i=edge[i].next)        {            v=edge[i].v,w=edge[i].w;            if(ll<=point[v].l&&point[v].l<=rr&&book[v]==0&&dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;                tem.v=v,tem.w=dis[v];                que.push(tem);            }        }    }    for(i=1;i<=n;i++)    {        ans=min(ans,dis[i]+point[i].p);    }    return;}int main(){    int u,v,w,t,p,l,r;    while(scanf("%d%d",&lim,&n)!=EOF)    {        memset(first,-1,sizeof(first));        num=0;        for(u=1;u<=n;u++)        {            scanf("%d%d%d",&point[u].p,&point[u].l,&t);            if(u==1) p=point[u].l;            while(t--)            {                scanf("%d%d",&v,&w);                addedge(u,v,w);            }        }        ans=point[1].p;        for(l=p-lim,r=p;l<=p;l++,r++)        {            calculate(l,r);        }        printf("%d\n",ans);    }    return 0;}

#include<stdio.h>
#include <queue>
#include <cstring>
#include <algorithm>
using namespacestd;

struct nodeI
{
    int p;
    int l;
};

struct nodeII{ int v;
    int w;
    int next;
};

struct nodeIII{ friendbool operator < (nodeIII n1,nodeIII n2)
    {
        return n1.w>n2.w;
    }
    int v;
    int w;
};

nodeI point[110];
nodeII edge[10010];
priority_queue <nodeIII> que;
int first[110],dis[110],book[110];
int lim,n,num,ans;

void addedge(int u,int v,int w){
    edge[num].v=v;
    edge[num].w=w;
    edge[num].next=first[u];
    first[u]=num++;
    return;
}

void calculate(int ll,int rr){
    nodeIII cur,tem;
    int i,u,v,w;
    while(!que.empty()) que.pop();
    memset(dis,0x3f,sizeof(dis));
    memset(book,0,sizeof(book));
    tem.v=1,tem.w=0;
    que.push(tem);
    dis[1]=0;
    while(!que.empty())
    {
        cur=que.top();
        que.pop();
        u=cur.v;
        if(book[u]==1)continue;
        book[u]=1;
        for(i=first[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v,w=edge[i].w;
            if(ll<=point[v].l&&point[v].l<=rr&&book[v]==0&&dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                tem.v=v,tem.w=dis[v];
                que.push(tem);
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        ans=min(ans,dis[i]+point[i].p);
    }
    return;
}

int main(){
    int u,v,w,t,p,l,r;
    while(scanf("%d%d",&lim,&n)!=EOF)
    {
        memset(first,-1,sizeof(first));
        num=0;
        for(u=1;u<=n;u++)
        {
            scanf("%d%d%d",&point[u].p,&point[u].l,&t);
            if(u==1) p=point[u].l;
            while(t--)
            {
                scanf("%d%d",&v,&w);
                addedge(u,v,w);
            }
        }
        ans=point[1].p;
        for(l=p-lim,r=p;l<=p;l++,r++)
        {
            calculate(l,r);
        }
        printf("%d\n",ans);
    }
    return 0;
}

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