UVA 11374 Airport Express （最短路）

题目只有一条路径会发生改变。

常见的思路，预处理出S和T的两个单源最短路，然后枚举商业线，商业线两端一定是选择到s和t的最短路。

路径输出可以在求最短路的同时保存pa数组得到一棵最短路树，也可以用dist数组检查。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 502, maxm = 2002;

int head[maxn], to[maxm], nxt[maxm],wei[maxm],ecnt;

void addEdge(int u,int v,int w)
{
    to[ecnt] = v;
    nxt[ecnt] = head[u];
    wei[ecnt] = w;
    head[u] = ecnt++;
}

typedef pair<int,int> Node;
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
int d1[maxn],d2[maxn];

void dijkstra(int s,int t,int (&d)[maxn])
{
    priority_queue<Node,vector<Node>,greater<Node> > q;
    memset(d,0x3f,sizeof(d));
    d[s] = 0; q.push(Node(0,s));
    while(q.size()){
        Node x = q.top(); q.pop();
        int u = x.se;
        if(x.fi != d[u]) continue;
        for(int i = head[u]; ~i; i = nxt[i]){
            int v = to[i];
            if(d[v] > d[u]+wei[i]){
                d[v] = d[u]+wei[i];
                q.push(Node(d[v],v));
            }
        }
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int N,S,E;
    bool first = true;
    while(~scanf("%d%d%d",&N,&S,&E)){
        if(!first) putchar('
');
        first = false;
        int M; scanf("%d",&M);
        memset(head,-1,sizeof(head));
        ecnt = 0;
        while(M--){
            int u,v,w; scanf("%d%d%d",&u,&v,&w);
            addEdge(--u,--v,w); addEdge(v,u,w);
        }
        dijkstra(--S,--E,d1);
        dijkstra(E,S,d2);

        int pick = -1, p2, ans = d1[E];
        int K; scanf("%d",&K);
        for(int i = 0; i < K; i++){
            int u,v,w; scanf("%d%d%d",&u,&v,&w);
            if(d1[--u] + w  < ans - d2[--v]){
                ans = d1[u] + w + d2[v];
                pick = u; p2 = v;
            }else if( d2[u] + w  < ans - d1[v]){
                ans = d2[u] + w + d1[v];
                pick = v; p2 = u;
            }
        }
        int u;
        if(~pick){
            stack<int> stk;
            u = pick;
            stk.push(u);
            while(u != S){
                for(int i = head[u]; ~i; i = nxt[i]){
                    if(d1[u] - wei[i] == d1[to[i]] ){
                        u = to[i]; stk.push(u); break;
                    }
                }
            }
            while(stk.size()){
                printf("%d ",stk.top()+1); stk.pop();
            }
            u = p2;
        }else {
            u = S;
        }
        while(u != E){
            printf("%d ",u+1);
            for(int i = head[u]; ~i; i = nxt[i]){
                if(d2[u] - wei[i] == d2[to[i]] ){
                    u = to[i];  break;
                }
            }
        }
        printf("%d
",E+1);
        if(~pick) printf("%d
",pick+1);
        else puts("Ticket Not Used");
        printf("%d
",ans);
    }
    return 0;
}