[HDU] 1224 Free DIY Tour 使用BELLMANFORD处理有向无回路图

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1224

方法：根据输入数据建立好的图（图中每个顶点会纪录当前累计获取的有趣度和自己本身的有趣度）是不会有回路的有向图，所以可以是用拓扑排序，由于题目输入数据就定下了第一个定点是起点，所以代码中直接从第一个顶点开始扫描的顺序就是拓扑排序的顺序，程序中不需要在进行拓扑排序。按该顺序每访问一个点，就松弛其所有的边，并设置其在最长（短）路径中的前驱。

感想：通过聚集分析，时间复杂度是o(v+e).

代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
int const MAX =0x3f3f3f3f;
//int const MAX = 10005;
int cityCount=0;
int straughtFlightCount=0;
struct straughtFlight
{
    int destination;
    straughtFlight* nextFlight;
    int source; 
};
struct cityNode
{
    int intrestingPoint;
    straughtFlight* firstFlight;
    bool visisted;
    int currentInstPont;
    cityNode* pre;
    int index;
};
cityNode* cites[102];
void createStraightFlight(int x,int y)
{
    straughtFlight* flight = (straughtFlight*)malloc(sizeof(straughtFlight));
    flight->destination=y;
    if(cites[x]->firstFlight == NULL)
        flight->nextFlight = NULL;
    else
        flight->nextFlight = cites[x]->firstFlight;
    cites[x]->firstFlight = flight;
}
struct cmp
{
    bool operator()(cityNode* x,cityNode* y)
    {
        if(x->currentInstPont<y->currentInstPont)
            return true;
        return false;
    }
};
int getTheMax(int startPos = 1)
{
    straughtFlight* t_flight;
    for(int i=startPos;i<=cityCount;i++)
    {
        t_flight = cites[i]->firstFlight;
        while(t_flight!=NULL)
        {
            if(cites[t_flight->destination]->intrestingPoint + cites[i]->currentInstPont > cites[t_flight->destination]->currentInstPont)
            {
                 cites[t_flight->destination]->currentInstPont = cites[t_flight->destination]->intrestingPoint + cites[i]->currentInstPont;
                 cites[t_flight->destination]->pre = cites[i];
            }
            t_flight=t_flight->nextFlight;
        }
    }
    return  cites[cityCount+1]->currentInstPont;
}
 
int main()
{
    int tc=0,t;
    scanf("%d",&t);
    while(tc<t)
    {
     
        scanf("%d",&cityCount);
        for(int i=1;i<=cityCount;i++)
        {
            cites[i] = (cityNode*)malloc(sizeof(cityNode));
            scanf("%d",&cites[i]->intrestingPoint);
            cites[i]->visisted=false;
            cites[i]->currentInstPont = -MAX;
            cites[i]->firstFlight = NULL;
            cites[i]->pre=NULL;
            cites[i]->index = i;
        }
        cites[1]->currentInstPont = 0;
        cites[cityCount+1]  = (cityNode*)malloc(sizeof(cityNode));
        cites[cityCount+1]->currentInstPont =-MAX;
        cites[cityCount+1]->intrestingPoint = 0;
        cites[cityCount+1]->firstFlight=NULL;
        cites[cityCount+1]->visisted = false;
        cites[cityCount+1]->pre=NULL;
        cites[cityCount+1]->index = 1;
        scanf("%d",&straughtFlightCount);
        int a,b;
        for(int i =0;i<straughtFlightCount;i++)
        {
            scanf("%d %d",&a,&b);
            if(a>b)
            {
                int t = a;
                a = b;
                b = t;    
            }
            createStraightFlight(a , b );
        }
 
          int re = getTheMax( 1);
        cityNode* t_n = cites[cityCount+1];
        stack<cityNode*> stk;
        while(t_n!=NULL)
        {
            stk.push(t_n);
            t_n = t_n->pre;
        }
        cout<<"CASE "<<tc+1<<"#"<<endl;
        cout<<"points : "<< (re<0 ? 0 :re) <<endl;
        cout<<"circuit : ";
        
        bool first = true;
        while(!stk.empty())
        {
            if(first)
            {
                first=false;
            }
            else
            {
                cout<<"->";
            }
            cout<<stk.top()->index;
            stk.pop();
        }
        cout<<endl;
        if(tc<t-1)
            cout<<endl;
        tc++;
    }
 
    return 0;
}