题意：给你一张无向图，t个可能的目的地，问在这t个点中哪些点的最短路中经过了g和h
思路：这是傻逼题，我直接dijstra用vector< set<int> > 保存路径， 最后再去判断下。。好像这并不是出题者想要的解法，不过时间还可以，300+ms
1 #pragma comment(linker, "/STACK:1000000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <fstream> 5 #include <algorithm> 6 #include <cmath> 7 #include <deque> 8 #include <vector> 9 #include <queue> 10 #include <string> 11 #include <cstring> 12 #include <map> 13 #include <stack> 14 #include <set> 15 #define LL long long 16 #define MAXN 100005 17 #define INF 0x3f3f3f3f 18 #define eps 1e-8 19 using namespace std; 20 struct Edge 21 { 22 int from, to, dist; 23 Edge(int from, int to, int dist):from(from), to(to), dist(dist){}; 24 }; 25 struct HeapNode 26 { 27 int d, u; 28 HeapNode(int d, int u):d(d), u(u){}; 29 bool operator <(const HeapNode& rhs) const{ 30 return d > rhs.d; 31 } 32 }; 33 vector< set<int> > road[MAXN]; 34 struct Dijstra 35 { 36 int n, m; 37 vector<Edge> edges; 38 vector<int> G[MAXN]; 39 bool done[MAXN]; 40 int d[MAXN]; 41 int p[MAXN]; 42 43 void init(int n){ 44 this->n = n; 45 for(int i = 0; i <= n; i++){ 46 G[i].clear(); 47 road[i].clear(); 48 } 49 edges.clear(); 50 } 51 52 void AddEdge(int from, int to, int dist){ 53 edges.push_back(Edge(from, to, dist)); 54 m = edges.size(); 55 G[from].push_back(m - 1); 56 } 57 58 void dijstra(int s){ 59 priority_queue<HeapNode> Q; 60 for(int i = 0; i <= n; i++){ 61 d[i] = INF; 62 } 63 d[s] = 0; 64 memset(done, 0, sizeof(done)); 65 Q.push(HeapNode(0, s)); 66 while(!Q.empty()){ 67 HeapNode x = Q.top(); 68 Q.pop(); 69 int u = x.u; 70 if(done[u]) continue; 71 done[u] = true; 72 for(int i = 0; i < G[u].size(); i++){ 73 Edge& e = edges[G[u][i]]; 74 if(d[e.to] > d[u] + e.dist){ 75 d[e.to] = d[u] + e.dist; 76 road[e.to].clear(); 77 if(road[u].empty()){ 78 set<int> tmp; 79 tmp.clear(); 80 tmp.insert(u); 81 road[e.to].push_back(tmp); 82 } 83 else{ 84 for(int j = 0; j < road[u].size(); j++){ 85 road[e.to].push_back(road[u][j]); 86 road[e.to][j].insert(u); 87 } 88 } 89 p[e.to] = G[u][i]; 90 Q.push(HeapNode(d[e.to], e.to)); 91 } 92 else if(d[e.to] == d[u] + e.dist){ 93 int w = road[e.to].size(); 94 for(int j = 0; j < road[u].size(); j++){ 95 road[e.to].push_back(road[u][j]); 96 road[e.to][w + j].insert(u); 97 } 98 } 99 } 100 } 101 } 102 }; 103 int n, m, t, g, h, s; 104 Dijstra p; 105 vector<int> res; 106 int main() 107 { 108 #ifndef ONLINE_JUDGE 109 freopen("in.txt", "r", stdin); 110 //freopen("out.txt", "w", stdout); 111 #endif // OPEN_FILE 112 int T; 113 scanf("%d", &T); 114 while(T--){ 115 scanf("%d%d%d", &n, &m, &t); 116 scanf("%d%d%d", &s, &g, &h); 117 int x, y, z; 118 p.init(n); 119 for(int i = 1; i <= m; i++){ 120 scanf("%d%d%d", &x, &y, &z); 121 p.AddEdge(x, y, z); 122 p.AddEdge(y, x, z); 123 } 124 p.dijstra(s); 125 res.clear(); 126 for(int i = 1; i <= t; i++){ 127 scanf("%d", &x); 128 for(int j = 0; j < road[x].size(); j++){ 129 road[x][j].insert(x); 130 if(road[x][j].find(g) != road[x][j].end() && road[x][j].find(h) != road[x][j].end()){ 131 res.push_back(x); 132 break; 133 } 134 } 135 } 136 sort(res.begin(), res.end()); 137 int w = res.size(); 138 for(int i = 0; i < w; i++){ 139 printf("%d ", res[i]); 140 } 141 printf(" "); 142 } 143 }