題目分析

我們可以從S出發演算法最短路，然後從E出發算出最短路，然後列舉每一條特殊通道即可。輸出可能有些麻煩。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4+100;
const int INF = 0x3f3f3f3f;

int N, S, E, M;
int d1[maxn], d2[maxn], vis[maxn];
int
 
 head[maxn], tot;

struct Edge{
    int to, next, val;
}e[maxn<<1];

void spfa(int S, int d[]){
    for(int i = 0; i < maxn; i++) d[i] = INF;
    memset(vis, 0, sizeof(vis));
    queue <int> q;
    q.push(S);
    vis[S] = 1;
    d[S] = 0;
    while(!q.empty()){
        int u = q.front(); q.pop();
        vis[u] = 0
 
;
        for(int i = head[u]; i != -1; i = e[i].next){
            int v = e[i].to;
            if(d[v] > d[u] + e[i].val){
                d[v] = d[u] + e[i].val;
                if(!vis[v]){
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}

void
 
 addedge(int from, int to, int val){
    e[tot].to = to;
    e[tot].val = val;
    e[tot].next = head[from];
    head[from] = tot++;
}

void init(){
    tot = 0;
    memset(head, -1, sizeof(head));
}

void print1(int u, int d[]){
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(d[u] == d[v] + e[i].val){
            print1(v, d);
            break;
        }
    }
    if(u == E) printf("%d\n", u);
    else printf("%d ", u);
}

void print2(int u, int d[]){
    if(u == E) printf("%d\n", u);
    else printf("%d ", u);
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(d[u] == d[v] + e[i].val){
            print2(v, d);
            break;
        }
    }
}

void solve(int f, int t){
    if(f == -1){
        print1(E, d1);
        printf("Ticket Not Used\n");
        return ;
    }
    print1(f, d1);
    print2(t, d2);
    printf("%d\n", f);
}

int main(){
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
    #endif // LOCAL
    int first = 0;
    while(scanf("%d%d%d", &N, &S, &E) != EOF){
        if(first) printf("\n");
        else first = 1;
        init();
        int from, to, val;
        scanf("%d", &M);
        for(int i = 0; i < M; i++){
            scanf("%d%d%d", &from, &to, &val);
            addedge(from, to, val);
            addedge(to, from, val);
        }
        spfa(S, d1);
        spfa(E, d2);
        scanf("%d", &M);
        int ans = d1[E];
        int f = -1, t = -1;
        for(int i = 0; i < M; i++){
            scanf("%d%d%d", &from, &to, &val);
            if(d1[from]+val+d2[to] < ans){
                ans = d1[from]+val+d2[to];
                f = from, t = to;
            }
            if(d1[to]+val+d2[from] < ans){
                ans = d1[to]+val+d2[from];
                f = to, t = from;
            }
        }
        solve(f, t);
        printf("%d\n", ans);
    }
    return 0;
}