[hdu P4114] Disney's FastPass

[hdu P4114] Disney's FastPass

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Disney's FastPass is a virtual queuing system created by the Walt Disney Company. First introduced in 1999 (thugh the idea of a ride reservation system was first introduced in world fairs), Fast-Pass allows guests to avoid long lines at the attractions on which the system is installed, freeing them to enjoy other attractions during their wait. The service is available at no additional charge to all park guests.
--- wikipedia

Disneyland is a large theme park with plenties of entertainment facilities, also with a large number of tourists. Normally, you need to wait for a long time before geting the chance to enjoy any of the attractions. The FastPass is a system allowing you to pick up FastPass-tickets in some specific position, and use them at the corresponding facility to avoid long lines. With the help of the FastPass System, one can arrange his/her trip more efficiently.
You are given the map of the whole park, and there are some attractions that you are interested in. How to visit all the interested attractions within the shortest time?

 

Input

The first line contains an integer T(1<=T<=25), indicating the number of test cases.
Each test case contains several lines.
The first line contains three integers N,M,K(1 <= N <= 50; 0 <= M <= N(N - 1)/2; 0 <= K <= 8), indicating the number of locations(starting with 1, and 1 is the only gate of the park where the trip must be started and ended), the number of roads and the number of interested attractions.
The following M lines each contains three integers A,B,D(1 <= A,B <= N; 0 <= D <= 10^4) which means it takes D minutes to travel between location A and location B.
The following K lines each contains several integers P_i, T_i, FT_i,N_i, F_i,1, F_i,2 ... F_i,Ni-1, F_iNi ,(1 <= P_i,N_i, F_i,j <=N, 0 <= FT_i <= T_i <= 10^4), which means the ith interested araction is placed at location Pi and there are Ni locations F_i,1; F_i,2 ... F_i,N_i where you can get the FastPass for the ith attraction. If you come to the ith attraction with its FastPass, you need to wait for only FTi minutes, otherwise you need to wait for Ti minutes.
You can assume that all the locations are connected and there is at most one road between any two locations.
Note that there might be several attrractions at one location.

 

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y is the minimum time of the trip.

Sample Input
2
4 5 2
1 2 8
2 3 4
3 4 19
4 1 6
2 4 7
2 25 18 1 3
4 12 6 1 3
4 6 2
1 2 5
1 4 4
3 1 1
3 2 1
3 4 1
2 4 10
2 8 3 1 4
4 8 3 1 2

Sample Output

Case #1: 53
Case #2: 14

Source
2011 Asia ChengDu Regional Contest

好题啊！只是刚开始一直被题意杀，样例都不知道为什么是这样。

好吧，那就解释一下题意吧。

一个人在一个城市里游玩，共n个景区，m条景区间相连的双向道路，还有k个景点，在n个景区里面（每个景区可能有多个或没有景点）。

然后给出一些景区间互达所需时间（就是边权），然后没有重边。

再给出每个景点的信息，包含：

这个景点在哪个景区里面，然后涉及到一个FastPass的问题。

就是说，如果你在游玩这个景点前有了这个景区的FastPass（注意是游玩而不是经过），那么只需要FTi的时间排队等候（就是游览自身所需时间），否则需要Ti的时间。

还有一点，就是FastPass相当于一张永久票，得到了就不会丢掉，并且同一时刻可以拿好几张。

所以样例二的解释就是，1-->3-->2(经过，并拿到4的票)-->4(游玩，并拿到2的票)-->3-->1。

总用时为(1)+(1)+(1+1+3)+(1+1+3)+(1+1)。

好，这下题意应该算是挺清楚了。我们来想做法。

在被题意杀之前我就写了个状压，然后GG了。

但每错，k那么小，显然是状压的料！

但首先，我们先要预处理出任意两个景区之间的最短路径，这是个贪心的思想，没有问题。

然后，我们设一个状态dp[s1][s2][i]。s1是已游览的点二进制集(0<=s1<1<<k)，s2是手上拥有的FastPass的点二进制集，i代表目前所在的点。

显然初始状态dp[0][0][1]=0，其他=inf。最终结果是dp[(1<<k)-1][i][1](0<=i<1<<k)。

然后我们来考虑方程。一般状压dp，用当前状态更新后继状态写的比较多，比较方便。

然后我们分两种情况：

1.下一个访问的景区同时要被参观景点-->dp[s1|(1<<j)][s2|own[b[j]]][b[j]]=min(dp[s1][s2][i])。

其中own[i]表示景区i拥有的某些景点的FastPass，这些节点组成二进制集，b[i]表示景点i所在的景区。

2.下一个访问的景区不被参观景点-->dp[s1][s2|own[j]][j]=min(dp[s1][s2][i])。

然后，算一下复杂度，O(T*(1<<8)*(1<<8)*(n^2))，像是要T的样子，但事实上没有满，测出来跑了2000ms，还可以。

code：

 %:pragma GCC optimize()
 #include<bits/stdc++.h>
 #define ms(a,x) memset(a,x,sizeof a)
 using namespace std;
 ,S=,inf=0x3c3c3c3c;
 int n,m,t,ans,f[N][N],b[N],t1[N],t2[N],cn[N],dp[S][S][N];
 inline int read() {
     ; char ch=getchar();
     ') ch=getchar();
     +ch-',ch=getchar();
     return x;
 }
 inline int bitcnt(int s) {
     ;
     )+bitcnt(s>>);
 }
 inline void upd(int &x,int y) {if (x>y) x=y;}
 int main() {
     ; cs<=T; cs++) {
         n=read(),m=read(),t=read(),ms(f,),ms(cn,);
         ,x,y,z; i<=m; i++) {
             x=read(),y=read(),z=read();
             z=min(z,f[x][y]),f[x][y]=f[y][x]=z;
         }
         ; i<=n; i++) f[i][i]=;
         ; k<=n; k++)
             ; i<=n; i++)
                 ; j<=n; j++) f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
         ,x,y; i<t; i++) {
             b[i]=read(),t1[i]=read(),t2[i]=read(),x=read();
             ; j<=x; j++) y=read(),cn[y]|=(<<i);
         }
         ms(dp,),dp[][][]=;
         ; s1<(<<t); s1++)
             ; s2<(<<t); s2++)
                 ; i<=n; i++) if (dp[s1][s2][i]<inf) {
                     ,c; j<t; j++) <<j))==) {
                         c=s2&(<<j)?t2[j]:t1[j],c+=f[i][b[j]];
                         upd(dp[s1|(<<j)][s2|cn[b[j]]][b[j]],dp[s1][s2][i]+c);
                     }
                     ; j<=n; j++)
                         upd(dp[s1][s2|cn[j]][j],dp[s1][s2][i]+f[i][j]);
                 }
         ans=inf;
         ; i<(<<t); i++) upd(ans,dp[(<<t)-][i][]);
         printf("Case #%d: %d\n",cs,ans);
     }
     ;
 }