poj 3635 带花费的Dij+head优化

练习！！

这里主要需要注意的是进队的条件和dp[][]状态的控制，dp[i][j]表示到第i个城市剩余汽油为j的最小花费。

代码：

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#define inf 100000000
#define MAXN 1200
#define MAXM 1200
using namespace std;

int head[MAXN],price[MAXN],dp[MAXN][120],vis[MAXN][120];

int temp,n,m;

struct Node{
    int v,len,next;
}edge[2*MAXN*10];

struct Node1{
    int d;
    int u;
    int cost;
    Node1(int x,int y,int z)
    {
        u=x;
        d=y;
        cost=z;
    }
    friend bool operator < (Node1 a,Node1 b)
    {
        return a.cost>b.cost;
    }
};
void addEdge(int x ,int y ,int c)
{
    edge[temp].v=x;
    edge[temp].len=c;
    edge[temp].next=head[y];
    head[y]=temp;
    temp++;

    edge[temp].v=y;
    edge[temp].len=c;
    edge[temp].next=head[x];
    head[x]=temp;
    temp++;
}

priority_queue<Node1> que;

int Dijstra(int s,int t,int c)
{
    while(!que.empty())que.pop();
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=c;j++)
        {
            dp[i][j]=inf;
        }
    }
    memset(vis,0,sizeof(vis));
    dp[s][0]=0;
    que.push(Node1(s,0,0));

    while(!que.empty())
    {
        Node1 b=que.top();
        que.pop();
        int u=b.u;
        int d=b.d;
        int cost=b.cost;
        vis[u][d]=1;
        if(u==t) return cost;

        if(d+1<=c&&!vis[u][d+1]&&dp[u][d+1]>dp[u][d]+price[u])
        {
            dp[u][d+1]=dp[u][d]+price[u];
            que.push(Node1(u,d+1,dp[u][d+1]));
        }
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v;
            v=edge[i].v;
            int len=edge[i].len;
            if(d>=len&&!vis[v][d-len]&&dp[v][d-len]>cost)
            {
                dp[v][d-len]=cost;
                que.push(Node1(v,d-len,cost));
            }
        }
    }
    return -1;
}

int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{

    for(int i=0;i<n;i++)
    {
        scanf("%d",&price[i]);
    }
    memset(head,-1,sizeof(head));
    temp=0;
    for(int i=1;i<=m;i++)
    {
        int v,w,c;
        scanf("%d%d%d",&v,&w,&c);
        addEdge(v,w,c);
    }

    int q;
    scanf("%d",&q);

    while(q--)
    {
        int q1,q2,q3;
        scanf("%d%d%d",&q1,&q2,&q3);
        int ans=Dijstra(q2,q3,q1);
        if(ans!=-1)
        {
            printf("%d\n",ans);
        }
        else
            printf("impossible\n");
    }
}
    return 0;
}