UVA 11374 Airport Express 最短路

最新推荐文章于 2021-09-07 23:12:41 发布

转载最新推荐文章于 2021-09-07 23:12:41 发布 · 96 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/BMan/p/3632895.html

本文介绍了一种结合商务票信息的最短路径算法实现。通过分别计算起点到各点及各点到终点的最短距离，并利用这些信息来确定是否使用特定商务票以达到整体路径成本最小化。

分别求出以S和E为起点的最短路，然后枚举每一张商务票 (u,v) 求 A(u) + w(u,v) + B(v)

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in1.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

const int maxn=505;

struct Edge
{
    int u,v,dist;
};
struct HeapNode
{
    int d,u;
    bool operator < (const HeapNode &ant) const
    {
        if(d!=ant.d)return d>ant.d;
        else return u>ant.u;
    }
};
struct dijksta
{
    int n;
    vector<int> g[maxn];
    vector<Edge> edge;
    int done[maxn];
    void init(int n)
    {
        this->n=n;
        for(int i=1;i<=n;i++)
            g[i].clear();
        edge.clear();
    }
    void add(int u,int v,int dist)
    {
        Edge e=(Edge){u,v,dist};
        edge.push_back(e);
        g[u].push_back(edge.size()-1);
    }

    void solve(int *d,int *p,int s)
    {
        priority_queue<HeapNode> q;
        memset(done,0,sizeof(done));
        for(int i=1;i<=n;i++)
            d[i]=INF;
        d[s]=0;
        p[s]=-1;
        q.push((HeapNode){0,s});
        while(!q.empty())
        {
            HeapNode x=q.top();q.pop();
            int u=x.u;
            if(done[u])continue;
            done[u]=1;
            for(int i=0;i<g[u].size();i++)
            {
                int v=edge[g[u][i]].v;
                int w=edge[g[u][i]].dist;
                if(w+d[u]<d[v])
                {
                    d[v]=d[u]+w;
                    q.push((HeapNode){d[v],v});
                    p[v]=u;
                }
            }
        }
    }
}solver;


int f[maxn],g[maxn],pf[maxn],pg[maxn];
Edge Ticket[maxn*4];

int check(Edge &e)
{
    return f[e.u]+e.dist+g[e.v];
}
int main()
{
    int n,s,e;
    int ca=0;
    while(scanf("%d%d%d",&n,&s,&e)!=EOF)
    {
        
        solver.init(n);
        int m;
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            solver.add(u,v,w);
            solver.add(v,u,w);
        }
        solver.solve(f,pf,s);
        solver.solve(g,pg,e);
        int k;
        scanf("%d",&k);
        for(int i=0;i<k;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            Ticket[i*2]=(Edge){u,v,w};
            Ticket[i*2+1]=(Edge){v,u,w};
        }

        int mint=f[e],minid=-1;
        for(int i=0;i<2*k;i++)
        {
            int x=check(Ticket[i]);
            if(x<mint){mint=x;minid=i;}
        }
        if(ca++!=0)printf("\n");
        if(minid<0)
        {
            for(int u=s;u!=-1;u=pg[u])
            {
                if(u!=s)printf(" ");
                printf("%d",u);
            }
            printf("\n");
            printf("Ticket Not Used\n%d\n",mint);
        }else
        {
            stack<int> s;
            Edge &e=Ticket[minid];
            for(int u=e.u;u!=-1;u=pf[u])
                s.push(u);
            while(!s.empty())
            {
                int u=s.top();s.pop();
                printf("%d",u);
                if(u!=e.u)printf(" ");
            }
            for(int u=e.v;u!=-1;u=pg[u])
                printf(" %d",u);
            printf("\n");
            printf("%d\n%d\n",e.u,mint);
        }
    }
    return 0;
}