hdu 4784 Dinner Coming Soon

            spfa+优先队列。刚开始只用的spfa，结果tle到死。然后听队友说要用到优先队列，想了想，对时间分层的话的确每一个结点都只进队列一次即可，因为只有大时间才能更新出小时间，然后就wa成shi了。按队友写的改了才过得，好伤心的说，这是好题。。。

            附上代码供大家对拍吧。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#define LL long long
#define CLR(a, b) memset(a, b, sizeof(a))
#define REP(i, n) for(int i = 0; i < n; i ++)
#define FF(i, a, b) for(int i = a; i < b; i ++)
#define FD(i, a, b) for(int i = a; i >= b; i --)
#define swp(a, b) a^=b^=a^=b;

using namespace std;

struct Node
{
    int u, b, vd, t;
    Node(){}
    Node(int a, int b, int c, int d):u(a), b(b), vd(c), t(d){}
    bool operator < (const Node &rhs) const
    {
        return t < rhs.t;
    }
};

struct Edge
{
    int u, v, t, m;
    Edge(){}
    Edge(int a, int b, int c, int d):u(a), v(b), t(c), m(d){}
}E[420];

int fir[220], next[440], tot;

void Add_Edge(int u, int v, int t, int m)
{
    E[tot] = Edge(u, v, t, m);
    next[tot] = fir[u]; fir[u] = tot ++;
}

int dp[110][6][7][220];
bool vis[110][6][7][220];
int p[110][7];
int N, M, B, K, R, T;

priority_queue<Node>q;

void relax(int u, int b, int vd, int t, int val)
{
    if ((u == 1 || u == N) && vd != 0) return ;
    if(dp[u][b][vd][t] < val)
    {
        dp[u][b][vd][t] = val;
        if(!vis[u][b][vd][t])
        {
            Node w = Node(u, b, vd, t);
            q.push(w);
            vis[u][b][vd][t] = true;
        }
    }
}

void spfa()
{
    int u, v, m, t,vd, b;
    Node a = Node(1, 0, 0, T);
    CLR(dp, -1);CLR(vis, false);
    dp[1][0][0][T] = R;
    vis[1][0][0][T] = true;
    q.push(a);
    while(!q.empty())
    {
        a = q.top();q.pop();
        u = a.u;b = a.b;vd = a.vd;t = a.t;
        if(u == N) continue;
        if(t > 0)
        {
            int val = dp[u][b][vd][t];
            relax(u, b, (vd + 1) % K, t - 1, val);
            if(p[u][vd] != -1)
            {
                if(b > 0)
                {
                    relax(u, b - 1, (vd + 1) % K, t - 1, val + p[u][vd]);
                }
                if(b < B && val >= p[u][vd])
                {
                    relax(u, b + 1, (vd + 1) % K, t - 1, val - p[u][vd]);
                }
            }
        }
        for(int i = fir[u]; ~i; i = next[i])
        {
            int tmp = t - E[i].t;
            if(tmp < 0) continue;
            v = E[i].v;
            int val = dp[u][b][vd][t] - E[i].m;
            if(val < 0) continue;
            relax(v, b, vd, tmp, val);
            if(p[u][vd] != -1)
            {
                if(b > 0)
                {
                    relax(v, b - 1, vd, tmp, val + p[u][vd]);
                }
                if(b < B && val >= p[u][vd])
                {
                    relax(v, b + 1, vd, tmp, val - p[u][vd]);
                }
            }
        }
    }
}

int main()
{
    int Time, cas = 1;
    scanf("%d", &Time);
    while(Time --)
    {
        scanf("%d%d%d%d%d%d", &N, &M, &B, &K, &R, &T);
        for(int i = 0; i < K; i ++)
        {
            for(int j = 1; j <= N; j ++)
            {
                scanf("%d", &p[j][i]);
            }
        }
        CLR(fir, -1);tot = 0;
        int u, v, t, m;
        for(int i = 0; i < M; i ++)
        {
            scanf("%d%d%d%d", &u, &v, &t, &m);
            Add_Edge(u, v, t, m);
        }
        spfa();
        int ans = -1;
        for(int i = 0; i <= B; i ++)
        {
            for(int j = 0; j <= T; j ++)
            {
                ans = max(ans, dp[N][i][0][j]);
            }
        }
        if(ans != -1) printf("Case #%d: %d
", cas ++, ans);
        else printf("Case #%d: Forever Alone
", cas ++);
    }
}