Coach Pang loves his boyfriend Uncle Yang very much. Today is Uncle Yang’s birthday, Coach Pang wants to have a romantic candlelit dinner at Uncle Yang’s house and he has to arrive there in T minutes.
There are N houses in their city numbered from 1 to N. Coach Pang lives in house 1 while Uncle Yang lives in house N. The houses are connected byM directed roads. It takes some time and usually a fee to pass one road. Coach Pang wants to reach Uncle Yang’s house before the dinner starts with as much money as possible.
But the matter is not so simple. Coach Pang decides to do some salt trade on the way to Uncle Yang’s house. The host of each house offers a price of a bag of salt, so Coach Pang can make a profit from the price differences. Each time when Coach Pang arrives at a house (except the house 1 and the house N). He can buy one bag of salt, sell one bag of salt or do nothing. Coach Pang can carry at most B bags of salt with him, and he carries no salt when he leaves his house. The trading is so efficient that the time cost of trading can be ignored.
However, the problem is more complicated than imagine. Coach Pang has a handheld device that can perform a journey around K parallel universes numbered from 0 to K-1. Coach Pang lives in the universe 0. When Coach Pang uses the device in universe i, he will be transported to the same place and the same time of universe (i+1) modK. The host of the house at the same place in different universe may offer a different price of salt. Luckily, the time cost and fee of the city roads are uniform among the K universes. The journey between universes costs no time but Coach Pang has to stand still watching the ads on the device for one minute every time before the device works. Remember, Coach Pang should never visit house 1 or house N in a universe other than universe 0, because the situation might become uncontrollable if he bumps into himself or his boyfriend in another universe.
The time is running out. Coach Pang asks you to tell him whether he can arrive at Uncle Yang’s house in time, and how much money Coach Pang can have at most when the dinner starts. Coach Pang has R yuan at the start, and will end his journey immediately once he arrives at Uncle Yang’s house. He must arrive at Uncle Yang’s house in T minutes, and he can’t have negative amount of money anywhere anytime. Please help him!
Input The first line of the input is an integer C representing the number of test cases.
For each test case, the first line will contain 6 integers N, M, B, K, R, T, as described above.
(2 <= N <= 100, 0 <= M <= 200, 1 <= B <= 4, 2 <= K <= 5, 0 <= R <= 10 5, 0 <= T <= 200)
The following K lines contain N integers each, indicating the price p ij (0 <= i < K, 1 <= j <= N) for a bag of salt offered by the host of house j in the universe i. The price of house 1 and house N will be marked as -1.(1 <= p ij <= 100)
Then M lines follow, each contains 4 integers a, b, t and m, indicating that there is a road from house a to house b that costs t minutes of time and m yuan of money. (1 <= a,b <= N, a<> b, 1 <= t <=15, 0 <= m <= 100)
Output For each test case, output one line containing “Case #x: y”, where x is the case number (starting from 1) and y is the most money Coach Pang can have if he can have dinner with Uncle Yang on time.
Print "Forever Alone" otherwise.Sample Input
2 3 2 1 2 10 6 -1 1 -1 -1 5 -1 1 2 1 0 2 3 1 1 2 2 1 2 5 5 -1 -1 -1 -1 1 2 10 2 1 2 2 10
Sample Output
Case #1: 17 Case #2: Forever Alone
对于这道题，只是想记录一下自己的代码……
1 #include <cstdio>
2 #include <iostream>
3 #include <algorithm>
4 #include <vector>
5 #include <set>
6 #include <map>
7 #include <string>
8 #include <cstring>
9 #include <stack>
10 #include <queue>
11 #include <cmath>
12 #include <ctime>
13 #include <utility>
14 using namespace std;
15 #define REP(I,N) for (I=0;I<N;I++)
16 #define rREP(I,N) for (I=N-1;I>=0;I--)
17 #define rep(I,S,N) for (I=S;I<N;I++)
18 #define rrep(I,S,N) for (I=N-1;I>=S;I--)
19 #define FOR(I,S,N) for (I=S;I<=N;I++)
20 #define rFOR(I,S,N) for (I=N;I>=S;I--)
21 typedef unsigned long long ull;
22 typedef long long ll;
23 const int INF=0x3f3f3f3f;
24 const ll INFF=0x3f3f3f3f3f3f3f3fll;
25 const ll M=1e6+3;
26 const ll maxn=1e5+7;
27 const int MAX=5e5+5;
28 const int MAX_N=MAX;
29 const ll MOD=1e6+3;
30 const double eps=0.00000001;
31 ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
32 template<typename T>inline T abs(T a) {return a>0?a:-a;}
33 inline ll powMM(ll a,ll b){ll ret=1;while (b){if (b&1)ret=ret*a%M;b>>=1;a=a*a%M;}return ret;}
34 int dp[101][202][10][10];
35 bool vi[101][202][10][10];
36 int head[205];//链表的起始点 储存从某个点出发的连向其他点的边的链表
37 int edgenum;//边的个数 从0开始
38 int prices[10][105];//盐的价格
39 bool arrival;//能否在T时间内到达n
40 struct edge
41 {
42 int target,t,money;
43 int nxt;
44 }edges[MAX];
45 void addedge(int now,int to,int time_cost,int money_cost)
46 {
47 edgenum++;
48 edges[edgenum].target=to;
49 edges[edgenum].t=time_cost;
50 edges[edgenum].money=money_cost;
51 edges[edgenum].nxt=head[now];
52 head[now]=edgenum;
53 // printf("now=%d edgenum=%d\n",now,edgenum);
54 }
55 struct node
56 {
57 int lo,t,k,pack;
58 bool operator <(const node &a)const
59 {
60 return t>a.t;
61 }
62 };
63 priority_queue<node> que;
64 void init()//初始化函数
65 {
66 memset(head,0xff,sizeof(head));//将链表头初始化为-1
67 memset(vi,false,sizeof(vi));
68 memset(dp,0,sizeof(dp));
69 edgenum=-1;
70 arrival=false;
71 while(!que.empty())//清空优先队列
72 que.pop();
73 }
74 int n,m,b,k,r,T;
75 int solve()
76 {
77 node tem;
78 tem.lo=1;tem.t=0;tem.k=0;tem.pack=0;
79 que.push(tem);
80 vi[1][0][0][0]=true;
81 dp[1][0][0][0]=r;
82 while(!que.empty())
83 {
84 tem=que.top();
85 que.pop();
86 if(tem.t>T||tem.lo==n)continue;
87 // printf("%d! %d\n",tem.lo,head[tem.lo]);
88 for(int i=head[tem.lo];i!=-1;i=edges[i].nxt)
89 {
90 // printf("nxt %d!\n",edges[i].nxt);
91 int new_time,new_money;
92 new_time=tem.t+edges[i].t;
93 new_money=dp[tem.lo][tem.t][tem.k][tem.pack]-edges[i].money;
94 if(new_time>T||new_money<0)continue;
95 if(edges[i].target==n&&tem.k!=0)continue;
96 if(edges[i].target==n)arrival=true;
97 if(tem.lo!=1&&tem.lo!=n)//是可以进行盐买卖的点
98 {
99 //买一包盐
100 if(tem.pack<b&&new_money-prices[tem.k][tem.lo]>dp[edges[i].target][new_time][tem.k][tem.pack+1])
101 {
102 dp[edges[i].target][new_time][tem.k][tem.pack+1]=new_money-prices[tem.k][tem.lo];
103 if(!vi[edges[i].target][new_time][tem.k][tem.pack+1])
104 {
105 vi[edges[i].target][new_time][tem.k][tem.pack+1]=true;
106 node nxt;
107 nxt.k=tem.k;nxt.lo=edges[i].target;nxt.pack=tem.pack+1;nxt.t=new_time;
108 que.push(nxt);
109 }
110 }
111 //卖一包盐
112 if(tem.pack&&new_money+prices[tem.k][tem.lo]>dp[edges[i].target][new_time][tem.k][tem.pack-1])
113 {
114 dp[edges[i].target][new_time][tem.k][tem.pack-1]=new_money+prices[tem.k][tem.lo];
115 if(!vi[edges[i].target][new_time][tem.k][tem.pack-1])
116 {
117 vi[edges[i].target][new_time][tem.k][tem.pack-1]=true;
118 node nxt;
119 nxt.k=tem.k;nxt.lo=edges[i].target;nxt.pack=tem.pack-1;nxt.t=new_time;
120 que.push(nxt);
121 }
122 }
123 }
124 //不买不卖 与无法买卖的情况相同
125 if(new_money>dp[edges[i].target][new_time][tem.k][tem.pack])
126 {
127 dp[edges[i].target][new_time][tem.k][tem.pack]=new_money;
128 if(!vi[edges[i].target][new_time][tem.k][tem.pack])
129 {
130 vi[edges[i].target][new_time][tem.k][tem.pack]=true;
131 node nxt;
132 nxt.k=tem.k;nxt.lo=edges[i].target;nxt.pack=tem.pack;nxt.t=new_time;
133 que.push(nxt);
134 }
135 }
136 }
137 //再看看是不是要穿越……
138 if(tem.lo!=1&&tem.lo!=n)
139 {
140 int new_time=tem.t+1,new_k=(tem.k+1)%k;
141 int new_money=dp[tem.lo][tem.t][tem.k][tem.pack];
142 if(new_time>T)
143 continue;
144 //买一包盐
145 if(tem.pack<b&&new_money-prices[tem.k][tem.lo]>dp[tem.lo][new_time][new_k][tem.pack+1])
146 {
147 dp[tem.lo][new_time][new_k][tem.pack+1]=new_money-prices[tem.k][tem.lo];
148 if(!vi[tem.lo][new_time][new_k][tem.pack+1])
149 {
150 vi[tem.lo][new_time][new_k][tem.pack+1]=true;
151 node nxt;
152 nxt.k=new_k;nxt.lo=tem.lo;nxt.pack=tem.pack+1;nxt.t=new_time;
153 que.push(nxt);
154 }
155 }
156 //卖一包盐
157 if(tem.pack&&new_money+prices[tem.k][tem.lo]>dp[tem.lo][new_time][new_k][tem.pack-1])
158 {
159 dp[tem.lo][new_time][new_k][tem.pack-1]=new_money+prices[tem.k][tem.lo];
160 if(!vi[tem.lo][new_time][new_k][tem.pack-1])
161 {
162 vi[tem.lo][new_time][new_k][tem.pack-1]=true;
163 node nxt;
164 nxt.k=new_k;nxt.lo=tem.lo;nxt.pack=tem.pack-1;nxt.t=new_time;
165 que.push(nxt);
166 }
167 }
168 //不买不卖
169 if(new_money>dp[tem.lo][new_time][new_k][tem.pack])
170 {
171 dp[tem.lo][new_time][new_k][tem.pack]=new_money;
172 if(!vi[tem.lo][new_time][new_k][tem.pack])
173 {
174 vi[tem.lo][new_time][new_k][tem.pack]=true;
175 node nxt;
176 nxt.k=new_k;nxt.lo=tem.lo;nxt.pack=tem.pack;nxt.t=new_time;
177 que.push(nxt);
178 }
179 }
180 }
181 }
182 if(!arrival)
183 return -1;
184 else
185 {
186 int an=0;
187 for(int i=0;i<=T;i++)
188 for(int j=0;j<=b;j++)
189 an=max(an,dp[n][i][0][j]);
190 return an;
191 }
192 }
193 int main()
194 {
195 int c;
196 scanf("%d",&c);
197 for(int Case=1;Case<=c;Case++)
198 {
199 init();
200 scanf("%d%d%d%d%d%d",&n,&m,&b,&k,&r,&T);
201 for(int i=0;i<k;i++)
202 for(int j=1;j<=n;j++)
203 scanf("%d",&prices[i][j]);//读入价格
204 for(int i=1;i<=m;i++)
205 {
206 int a,b,t,m;
207 scanf("%d%d%d%d",&a,&b,&t,&m);
208 addedge(a,b,t,m);
209 }
210 int an=solve();
211 if(an==-1)
212 printf("Case #%d: Forever Alone\n",Case);
213 else
214 printf("Case #%d: %d\n",Case,an);
215 }
216 return 0;
217 }