UVA 11374 - Airport Express（最短路）

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题目链接：点击打开链接

题意：某个人要从起点到终点，有m个经济站和k个商业站，只能乘坐一次商业站，求最短时间。

思路：因为多了商业站，所以不能直接套用最短路，但是注意到只有一次乘坐商业站的机会，所以直接枚举坐哪个商业站就行了，这样，其他部分就是最短路了，假设商业站是从a到b，花费c，那么答案就是d1[a] + d2[b] + c， d1是从起点的最短路，d2是从终点出发的最短路。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 500 + 10;int T,n,m,S,d1[maxn], p1[maxn], d2[maxn], p2[maxn], vis[maxn], ok, dist;struct node {    int u, val;    node(int u=0, int val=0):u(u), val(val) {}    bool operator < (const node& rhs) const {        return val > rhs.val;    }};void print(int root, int p[], int id, int S) {    vector<int> ans;    while(root != S) {        ans.push_back(root);        root = p[root];    }    ans.push_back(root);    int len = ans.size();    if(id == 1) for(int i=len-1;i>=0;i--) {        if(i != len-1) printf(" ");        printf("%d",ans[i]);    }    else for(int i=0;i<len;i++) {        if(i != 0) printf(" ");        printf("%d",ans[i]);    }}vector<node> g[maxn];void BFS(int haha, int d[], int p[]) {    priority_queue<node> q;    q.push(node(haha, 0));    for(int i=1;i<=n;i++) {        d[i] = INF;    }    d[haha] = 0;    memset(vis, false, sizeof(vis));    while(!q.empty()) {        node u = q.top(); q.pop();        if(vis[u.u]) continue;        vis[u.u] = true;        int len = g[u.u].size();        for(int i=0;i<len;i++) {            node v = g[u.u][i];            if(d[v.u] > d[u.u] + v.val) {                d[v.u] = d[u.u] + v.val;                p[v.u] = u.u;                q.push(node(v.u, d[v.u]));            }        }    }}int a,b,c,kase=0;int main() {    while(~scanf("%d%d%d",&n,&S,&T)) {        scanf("%d",&m);        for(int i=1;i<=n;i++) g[i].clear();        while(m--) {            scanf("%d%d%d",&a,&b,&c);            g[a].push_back(node(b, c));            g[b].push_back(node(a, c));        }        scanf("%d",&m);        ok = -1;        BFS(S, d1, p1);        BFS(T, d2, p2);        int s = -1,t = -1,ans = d1[T], res = 0;        for(int i=0;i<m;i++) {            scanf("%d%d%d",&a,&b,&c);            int cur = d1[a] + d2[b] + c;            if(cur < ans) {                ans = cur;                s = a; t = b;            }            cur = d1[b] + d2[a] + c;            if(cur < ans) {                ans = cur;                s = b; t = a;            }        }        if(kase) printf("\n");        else ++kase;        if(s > 0) {            print(s, p1, 1, S); printf(" ");            print(t, p2, 2, T); printf("\n");            printf("%d\n",s);        }        else {            print(T, p1, 1, S); printf("\n");            printf("Ticket Not Used\n");        }        printf("%d\n",ans);    }    return 0;}

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