HDU1224 DP

Free DIY Tour

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6227    Accepted Submission(s): 2013

Problem Description

Weiwei
is a software engineer of ShiningSoft. He has just excellently
fulfilled a software project with his fellow workers. His boss is so
satisfied with their job that he decide to provide them a free tour
around the world. It's a good chance to relax themselves. To most of
them, it's the first time to go abroad so they decide to make a
collective tour.

The tour company shows them a new kind of tour
circuit - DIY circuit. Each circuit contains some cities which can be
selected by tourists themselves. According to the company's statistic,
each city has its own interesting point. For instance, Paris has its
interesting point of 90, New York has its interesting point of 70, ect.
Not any two cities in the world have straight flight so the tour company
provide a map to tell its tourists whether they can got a straight
flight between any two cities on the map. In order to fly back, the
company has made it impossible to make a circle-flight on the half way,
using the cities on the map. That is, they marked each city on the map
with one number, a city with higher number has no straight flight to a
city with lower number.

Note: Weiwei always starts from
Hangzhou(in this problem, we assume Hangzhou is always the first city
and also the last city, so we mark Hangzhou both 1 and N+1), and its interesting point is always 0.

Now as the leader of the team, Weiwei wants to make a tour as interesting as possible. If you were Weiwei, how did you DIY it?

 

Input

The input will contain several cases. The first line is an integer T which suggests the number of cases. Then T cases follows.
Each case will begin with an integer N(2 ≤ N ≤ 100) which is the number of cities on the map.
Then N integers follows, representing the interesting point list of the cities.
And
then it is an integer M followed by M pairs of integers [Ai, Bi] (1 ≤ i
≤ M). Each pair of [Ai, Bi] indicates that a straight flight is
available from City Ai to City Bi.

 

Output

For
each case, your task is to output the maximal summation of interesting
points Weiwei and his fellow workers can get through optimal DIYing and
the optimal circuit. The format is as the sample. You may assume that
there is only one optimal circuit.

Output a blank line between two cases.

 

Sample Input

2

3

0 70 90

4

1 2

1 3

2 4

3 4

3

0 90 70

4

1 2

1 3

2 4

3 4

 

Sample Output

CASE 1#

points : 90

circuit : 1->3->1

 

CASE 2#

points : 90

circuit : 1->2->1

 

Author

JGShining（极光炫影）

 

Source

杭州电子科技大学第三届程序设计大赛

 

题意：

有n个城市，每次只能从1城市出发最后回到1 城市，给出几个相连的城市并且只能从数字大的城市到数字小的城市即路是单向的，每个城市有一个价值，问怎么走一圈能够得到最多的价值。

代码：

 /*
 到达每一个城市的最大价值取决于能够到达他的城市中价值较大的那个再加上它本身，记录路径load[i]存入能够到达i
 的最优的那个城市最后输出load[i],load[load[i]]........
 */
 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<iomanip>
 #include<queue>
 #include<stack>
 using namespace std;
 int t,n,m;
 int dp[];
 int val[];
 int main()
 {
     int a,b;
     scanf("%d",&t);
     for(int h=;h<=t;h++)
     {
         stack<int>q[];
         scanf("%d",&n);
         for(int i=;i<=n;i++)
         scanf("%d",&val[i]);
         val[n+]=;
         scanf("%d",&m);
         while(m--)
         {
             scanf("%d%d",&a,&b);
             q[b].push(a);
         }
         memset(dp,,sizeof(dp));
         int load[];
         for(int i=;i<=n+;i++)
         {
             if(q[i].empty())
             continue;
             while(!q[i].empty())
             {
                 int father=q[i].top();
                 q[i].pop();
                 if(dp[i]<dp[father]+val[i])
                 {
                     dp[i]=dp[father]+val[i];
                     load[i]=father;
                 }
             }

         }
         if(h!=)
         printf("\n");
         printf("CASE %d#\n",h);
         printf("points : %d\ncircuit : ",dp[n+]);
         int k=,kk=n+,lload[];
         while(kk!=)
         {
             lload[k++]=load[kk];
             kk=load[kk];
         }
         for(int i=k-;i>=;i--)
         {
             printf("%d->",lload[i]);
         }
         printf("1\n");
     }
     return ;
 }