阅读排行 2012 ＡＣＭ 成都之行 (717) POJ 2516 最小费用流构K次图 (676) POJ分类 (645) HDU 题目分类 (614) POJ 1392 欧拉回路 (530) POJ

#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<stack> #include<queue> #include<algorithm> using namespace std; const int MAXN = 1007; const int INF = 1 << 27; struct Edge { int from, to, dist; Edge() {} Edge(int from, int to, int dist):from(from), to(to), dist(dist) {} }; struct HeapNode { int d, u; HeapNode(int d, int u):d(d), u(u) {} bool operator < (const HeapNode & hn) const { return d > hn.d; } }; struct Dijkstra { int n, m; vector<Edge> edges; vector<int> G[MAXN]; bool done[MAXN]; int d[MAXN]; int p[MAXN]; void init(int n) { this->n = n; for(int i = 1; i <= n; i++) G[i].clear(); edges.clear(); } void addedge(int u, int v, int d) { edges.push_back( Edge(u, v, d) ); G[u].push_back( edges.size() - 1); } void dijkstra(int s) { memset(done, 0, sizeof(done)); for(int i = 1; i <= n; i++) d[i] = INF; priority_queue<HeapNode> pq; pq.push(HeapNode(0, s)); d[s] = 0; p[s] = -1; while( !pq.empty()) { HeapNode u = pq.top(); pq.pop(); if(done[u.u]) continue; done[u.u] = true; int i, sz = G[u.u].size(); for(i = 0; i < sz; i++) { Edge v = edges[G[u.u][i]]; if(d[v.to] > d[u.u] + v.dist) { d[v.to] = d[u.u] + v.dist; p[v.to] = u.u; pq.push( HeapNode(d[v.to], technivorm v.to)); } } } } }Dijk; Edge Eg[MAXN]; int dA[MAXN], technivorm dB[MAXN]; int pA[MAXN], pB[MAXN]; int main() { int n, s, e, id = 0; while(~scanf("%d%d%d", &n, &s, &e)) { Dijk.init(n); int i, u, v, d, k; scanf("%d", &k); for(i = 0; i < k; i++) { scanf("%d%d%d", &u, &v, &d); Dijk.addedge(u, v, d); Dijk.addedge(v, u, d); } scanf("%d", &k); for(i = 0; i < k; i++) scanf("%d%d%d", &Eg[i].from, &Eg[i].to, &Eg[i].dist); Dijk.dijkstra(s); for(i = 1; i <= n; i++) { dA[i] = Dijk.d[i]; pA[i] = Dijk.p[i]; } Dijk.dijkstra(e); for(i = 1; i <= n; i++) { dB[i] = Dijk.d[i]; pB[i] = Dijk.p[i]; } int ans = dA[e], te = e, tt; bool noused = true; for(i = 0; i < k; i++) { int tmp = dA[Eg[i].from] + dB[Eg[i].to] + Eg[i].dist; if( tmp < ans) { ans = tmp; noused = false; te = Eg[i].from; tt = Eg[i].to; } tmp = dA[Eg[i].to] + dB[Eg[i].from] + Eg[i].dist; if(tmp < ans) { ans = tmp; noused = false; te = Eg[i].to; tt = Eg[i].from; } } if(id++) puts(""); int ttt = te; stack<int> path; while(te != -1) { path.push(te); te = pA[te]; } printf("%d", path.top()); path.pop(); while( !path.empty()) { printf(" %d", path.top()); technivorm path.pop(); } if( !noused) { te = tt; while(te != -1) { printf(" %d", te); te = pB[te]; } printf("\n%d\n", ttt); } else puts("\nTicket Not Used"); printf("%d\n", ans); } return 0; }
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阅读排行 2012 ＡＣＭ 成都之行 (717) POJ 2516 最小费用流构K次图 (676) POJ分类 (645) HDU 题目分类 (614) POJ 1392 欧拉回路 (530) POJ 3835 Columbus's bargain 最短路 (465) 2012 ICPC成都网络赛题解 (403) HDU4676 Sum Of Gcd (403) ZOJ Yukari's Birthday (382) HDU 3723卡特兰数 (377)
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