UVA 11374 Airport Express（枚举+最短路）

枚举每条商业线<a, b>，设d[i]为起始点到每点的最短路，g[i]为终点到每点的最短路，ans便是min{d[a] + t[a, b] + g[b]}。注意下判断是否需要经过商业线。输出也有点坑的，每两组间用空行隔开。。。

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<fstream>
#include<sstream>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define LL long long
#define PB push_back
#define debug puts("**debug**")
using namespace std;

const int maxn = 1000;
const int INF = 10000000;
int n, s, e, m, k, x[maxn], y[maxn], z[maxn], flag;
struct Heap
{
    int d, u;
    bool operator < (const Heap& rhs) const
    {
        return d > rhs.d;
    }
};
struct Edge
{
    int from, to, dist;
};
vector<Edge> edges;
vector<int> G[maxn];
bool done[maxn];
int d[2][maxn], p[2][maxn], path[maxn];

inline void init()
{
    REP(i, n+1) G[i].clear(); edges.clear();
}

void dij(int s, int cur)
{
    priority_queue<Heap> q; q.push((Heap){0, s});
    REP(i, n+1) d[cur][i] = INF; d[cur][s] = 0;
    CLR(done, 0);
    while(!q.empty())
    {
        Heap x = q.top(); q.pop();
        int u = x.u, nc = G[u].size();
        if(done[u]) continue;
        done[u] = 1;
        REP(i, nc)
        {
            Edge& e = edges[G[u][i]];
            if(d[cur][e.to] > d[cur][u] + e.dist)
            {
                d[cur][e.to] = d[cur][u] + e.dist;
                p[cur][e.to] = edges[G[u][i]].from;
                q.push((Heap){d[cur][e.to], e.to});
            }
        }
    }
}

void add(int from, int to, int dist)
{
    edges.PB((Edge){from, to, dist});
    edges.PB((Edge){to, from, dist});
    int nc = edges.size();
    G[from].PB(nc-2);
    G[to].PB(nc-1);
}

void read()
{
    scanf("%d", &m);
    REP(i, m)
    {
        scanf("%d%d%d", &x[i], &y[i], &z[i]);
        add(x[i], y[i], z[i]);
    }
    scanf("%d", &k);
    REP(i, k) scanf("%d%d%d", &x[i], &y[i], &z[i]);
}

void solve()
{
    dij(s, 0);
    dij(e, 1);
    int ans = d[0][e], tot = d[0][e], a, b;
    REP(i, k)
    {
        int tmp = d[0][x[i]] + d[1][y[i]] + z[i];
        if(tmp < ans) ans = tmp, a = x[i], b = y[i];
        tmp = d[0][y[i]] + d[1][x[i]] + z[i];
        if(tmp < ans) ans = tmp, a = y[i], b = x[i];
    }
    if(ans == tot)
    {
        int cnt = 0, u = e; path[cnt++] = e;
        while(u != s) path[cnt++] = p[0][u], u = p[0][u];
        FD(i, cnt-1, 0) printf("%d%c", path[i], i == 0 ? '\n' : ' ');
        puts("Ticket Not Used");
        printf("%d\n", tot);
    }
    else
    {
        int cnt = 0, tmp[maxn], u = a;
        while(u != s) tmp[cnt++] = p[0][u], u = p[0][u];
        REP(i, cnt) path[i] = tmp[cnt-i-1];

        path[cnt] = a, path[++cnt] = b, cnt++;
        u = b;
        while(u != e) path[cnt++] = p[1][u], u = p[1][u];

        REP(i, cnt) printf("%d%c", path[i], i == cnt - 1 ? '\n' : ' ');
        printf("%d\n%d\n", a, ans);
    }
}

int main()
{
    flag = 0;
    while(~scanf("%d%d%d", &n, &s, &e))
    {
        if(flag) puts("");
        flag++;

        init();

        read();

        solve();
    }
    return 0;
}