HDU1535 Invitation Cards 堆优化
题意：从起点1到另外所有点的距离之和加上，另外所有点到1的距离之和，来去的花费不同。
两次Dijkstra，第二次反方向时，改变始末位置便可套用之前的Dijkstra模板。
1 #include<cstdio> 2 #include<queue> 3 #include<algorithm> 4 #include<vector> 5 #define maxn 1000010 6 using namespace std; 7 int n, m,ans = 0; 8 int a[maxn], b[maxn], c[maxn], dis[maxn]; 9 bool done[maxn]; 10 const int inf=0x3f3f3f3f; 11 struct edge { 12 int from, to, w; 13 }; 14 vector<edge>e[maxn]; 15 struct s_node { 16 int id, n_dis; 17 bool operator<(const s_node& a)const { return n_dis > a.n_dis; } 18 }; 19 void init() {for (int i = 1;i <= m;i++)dis[i] = inf, done[i] = false;} 20 void INIT() {for (int i = 1;i <= m;i++)e[i].clear();} 21 void dij() { 22 int s = 1; 23 init(); 24 dis[s] = 0; 25 priority_queue<s_node>Q; 26 Q.push( s_node{s,dis[s]} ); 27 while (!Q.empty()) { 28 s_node u = Q.top(); 29 Q.pop(); 30 if (done[u.id])continue; 31 done[u.id] = true; 32 int len = e[u.id].size(); 33 for (int i = 0;i < len;i++) { 34 edge y = e[u.id][i]; 35 if (done[y.to])continue; 36 if (dis[y.to] > y.w + u.n_dis) { 37 dis[y.to] = y.w + u.n_dis; 38 Q.push({ y.to,dis[y.to] }); 39 } 40 } 41 } 42 INIT(); 43 for (int i = 2;i <= n;i++)ans += dis[i]; 44 } 45 int main() 46 { 47 int T; 48 scanf("%d", &T); 49 while (T--) { 50 ans = 0; 51 scanf("%d%d", &n, &m); 52 for(int i=1;i<=m;i++)scanf("%d%d%d", &a[i], &b[i], &c[i]); 53 for(int i=1;i<=m;i++)e[a[i]].push_back(edge{ a[i],b[i],c[i] }); 54 dij(); 55 for (int i = 1;i <= m;i++)e[b[i]].push_back(edge{ b[i],a[i],c[i] }); 56 dij(); 57 printf("%d\n", ans); 58 } 59 }
