杭电1224（Floyd最长路+输出路径）

Free DIY Tour

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

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

思路：

题意：自定义旅游，每个城市 n 有不同的兴趣指数，起始城市 0 和终点城市 n+1为同一个且兴趣指数为0，在标号大的城市没有飞往标号小的城市的飞机（即单向图），求出最大的兴趣指数并输出路径。

方法：Floyd算法求最长路径。

注意：求最长路时，INF初始值为 -0xfffffff 。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 105
#define INF -0xfffffff //-的
int n,m;
int city[maxn];//记录城市的好玩程度 
int fun[maxn][maxn];//以连通边形式记录好玩程度 
int path[maxn][maxn];//记录路径 

void init()
{
    city[n]=0;//初始化第n+1个城市为起始城市且好玩程度为0 
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(i==j) fun[i][j]=0;
            else fun[i][j]=INF;
            path[i][j]=j;// 
        }
    }
}

void floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(fun[i][j]<fun[i][k]+fun[k][j])
                {
                    fun[i][j]=fun[i][k]+fun[k][j];
                    path[i][j]=path[i][k];//
                }
            }
        }
    }
}

int main()
{
    int t,u,v;//t case,城市u->v
    int icase=0;//记录case 
    cin>>t;
    while(t--)
    {
        cin>>n;//n个城市 
        n++;//第n+1个城市也为起始城市 
        init();
        for(int i=1;i<n;i++)
        {
            cin>>city[i];//输入城市有趣程度 
        } 
        cin>>m;//m对城市连通情况 
        for(int i=0;i<m;i++)
        {
            cin>>u>>v;
            fun[u][v]=city[v];//u->v有趣程度为v的有趣程度 
        }
        floyd();
        cout<<"CASE "<<++icase<<"#"<<endl;
        cout<<"points : "<<fun[1][n]<<endl;
        cout<<"circuit : 1";
        //输出路径信息 
        int i=1;
        while(i!=n)
        {
            if(path[i][n]==n) cout<<"->"<<1;
            else cout<<"->"<<path[i][n];
            i=path[i][n];
        }
        cout<<endl; 
        if(t!=0) cout<<endl;//两个输出间要有空行 
    }
    return 0;
}