UVA 11374 Airport Express（最短路）

最短路。

把题目抽象一下：已知一张图，边上的权值表示长度。现在又有一些边，只能从其中选一条加入原图，使起点->终点的距离最小。

当加上一条边a->b，如果这条边更新了最短路，那么起点st->终点ed的最小距离=st->a  +  a->b  +b->ed 三个值的最短距离之和。于是正反求两次单元最短路。再将k条边遍历一遍就好了。

最近在改代码风格，写起来很别扭。。uva又挂了，最让我不理解的是http://www.cnblogs.com/arbitrary/archive/2013/02/06/2908099.html这个家伙的代码竟然能ac，问题是我稍微改一下就submission error，难道代码还能防盗= =

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define clr(a,x) memset(a,x,sizeof(a));
#define ref(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

const int MAXN=1111;
const int INF =55555;

struct Edge{
    int u,v,c;
};

struct Node{
    int d,u;
    bool operator <(const Node x)const {
        return d>x.d;
    }
};

struct Dij{
    int n,m;
    vector<Edge>edge;
    vector<int>G[MAXN];
    bool vis[MAXN];
    int d[MAXN];
    int p[MAXN];

    void init(int n)
    {
        this->n=n;
        ref(i,0,n-1)G[i].clear();
        edge.clear();
    }

    void add(int u,int v,int c)
    {
        edge.push_back((Edge){u,v,c});
        m=edge.size();
        G[u].push_back(m-1);
    }

    void dij(int st)
    {
        priority_queue<Node>q;
        ref(i,0,n-1)d[i]=(i==st)?0:INF;
        clr(vis,0);
        q.push((Node){0,st});
        while(!q.empty())
        {
            Node x=q.top();
            q.pop();
            int u=x.u;
            if(vis[u])
                continue;
            vis[u]=true;
            ref(i,0,G[u].size()-1){
                Edge e=edge[G[u][i]];
                if(d[e.v]>d[u]+e.c){
                    d[e.v]=d[u]+e.c;
                    p[e.v]=G[u][i];
                    q.push((Node){d[e.v],e.v});
                }
            }
        }
    }
};

Dij solver;
int d1[MAXN],d2[MAXN];
int p[MAXN];

int main()
{
    int cnt=0;
    int n,m,k,st,ed;
    while(~scanf("%d%d%d",&n,&st,&ed))
    {
        if(cnt++)
            puts("");

        st--;ed--;
        solver.init(n);

        int u,v,c;
        scanf("%d",&m);
        ref(i,0,m-1){
            scanf("%d%d%d",&u,&v,&c);
            u--;v--;
            solver.add(u,v,c);
            solver.add(v,u,c);
        }

        solver.dij(st);
        ref(i,0,n-1)
            d1[i]=solver.d[i];

        solver.dij(ed);
        ref(i,0,n-1){
            d2[i]=solver.d[i];
            p[i]=solver.p[i];
        }
        p[ed]=-1;

        scanf("%d",&k);
        int s=d1[ed],a,b;
        a=b=-1;
        ref(i,0,k-1){
            scanf("%d%d%d",&u,&v,&c);
            u--;v--;
            if(s>d1[u]+c+d2[v]){
                s=d1[u]+c+d2[v];
                a=u;b=v;
            }
            if(s>d1[v]+c+d2[u]){
                s=d1[v]+c+d2[u];
                a=v;b=u;
            }
        }
        
        int flog=0;
        if(a==-1){       //不乘商务车
            int x=st;

            while(x!=ed)
            {
                if(!flog){printf("%d",x+1);flog=1;}
                else printf(" %d",x+1);
                x=solver.edge[p[x]].u;       //p[]中保存的是 u->v 这条边的序号
            }
            printf(" %d",ed+1);
            printf("
Ticket Not Used
%d
",s);
        }else{        //乘坐商务车
            int x=st;
            while(x!=a)
            {
                if(!flog){printf("%d",x+1);flog=1;}
                else printf(" %d",x+1);
                x=solver.edge[p[x]].u;
            }
            if(x!=st)printf(" %d",x+1);
            else printf("%d",x+1);
            x=b;
            while(x!=ed)
            {
                printf(" %d",x+1);
                x=solver.edge[p[x]].u;
            }
            printf(" %d
",ed+1);
            printf("%d
%d
",a+1,s);
        }
    }
    return 0;
}

/*
4 1 4
4
1 2 2
1 3 3
2 4 4
3 4 5
1
1 4 7

2 1 2
1
1 2 2
1
1 2 1
*/