传送门：
http://acm.hdu.edu.cn/showproblem.php?pid=1224
Free DIY Tour
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8692 Accepted Submission(s): 2804
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?
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.
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.
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
Recommend
分析：
题目意思：
维维要去旅游，然后旅游公司表示有DIY线路，维维可以自己选线路去旅游，每个城市都有一定的兴趣点，维维想要在这次旅游中获得最多的兴趣点数，另外起点在杭州，终点也一定是在杭州，只是中途的城市以及路线要自己DIY，并且绝对不会有从大编号城市到小编号城市的路线（城市编号从1到N，另外N+1表示终点），显示输入测试组数，然后每组测试先是输入一个城市数量N（2<=N<=100，包括起点），然后依次是N个城市的兴趣点数，再输入路线数量，接下来每条路输入两个数A和B，表示有从A到B的路线。最后你要按照格式输出获得的最大兴趣点以及路线就是了。
这里的终点为n+1，初始化时不要忘了[n+1]城市。同时输出时应将“n+1”改为1号城市
code：
#include <iostream> #include <cstdio> #include<stdio.h> #include<algorithm> #include<cstring> #include<math.h> #include<memory> #include<queue> using namespace std; typedef long long LL; #define max_v 120 int n,m; int G[max_v][max_v]; int a[max_v]; int dis[max_v];//保存价值 int way[max_v];//保存路 void spfa(int s,int tn) { for(int i=1;i<=tn;i++) { dis[i]=0; way[i]=0; } queue<int>q; q.push(s); int p; while(!q.empty()) { p=q.front(); q.pop(); for(int i=1;i<=tn;i++) { if(G[p][i]!=0) { if(dis[p]+a[i]>dis[i]) { dis[i]=dis[p]+a[i]; way[i]=p; q.push(i); } } } } } void pri(int tn)//路径打印 { int a[300]; int c=1; int i=tn; while(way[i]) { a[c++]=way[i]; i=way[i]; } for(int i=c-1;i>=1;i--) { printf("%d->",a[i]); } printf("1 "); } int main() { int t; scanf("%d",&t); int k=1; while(t--) { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } a[n+1]=0; scanf("%d",&m); memset(G,0,sizeof(G)); for(int i=1;i<=m;i++) { int x,y; scanf("%d %d",&x,&y); G[x][y]=1; } spfa(1,n+1); if(k!=1) printf(" "); printf("CASE %d# ",k++); printf("points : %d ",dis[n+1]); printf("circuit : "); pri(n+1); } return 0; }