水水更健康。。。话说回来，这真的是水题？T T
首先，容易想到：
令ans1 = t1为源，到s和t2的距离之和；ans2 = t2为源，到s和t1的距离之和
ans = min(ans1, ans2)
然后，开始写单元最短路。。。spfa。。。
1 /************************************************************** 2 Problem: 2100 3 User: rausen 4 Language: C++ 5 Result: Time_Limit_Exceed 6 ****************************************************************/ 7 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 100005; 15 const int M = 400005; 16 17 struct edges { 18 int next, to, v; 19 edges() {} 20 edges(int _next, int _to, int _v) : next(_next), to(_to), v(_v) {} 21 }e[M]; 22 23 int n, m, s, T1, T2; 24 int first[N], tot; 25 int q[N], l, r, dis[N]; 26 bool v[N]; 27 int ans; 28 29 inline int read() { 30 int x = 0; 31 char ch = getchar(); 32 while (!isdigit(ch)) 33 ch = getchar(); 34 while (isdigit(ch)) { 35 x = x * 10 + ch - '0'; 36 ch = getchar(); 37 } 38 return x; 39 } 40 41 inline void Add_Edges(int x, int y, int z) { 42 e[++tot] = edges(first[x], y, z), first[x] = tot; 43 e[++tot] = edges(first[y], x, z), first[y] = tot; 44 } 45 46 void spfa(int S) { 47 memset(dis, 127 / 3, sizeof(dis)); 48 q[0] = S, dis[S] = 0, v[S] = 1; 49 int p, x, y; 50 for (l = 0, r = 0; l != (r + 1) % N; (++l) %= N) { 51 p = q[l]; 52 for (x = first[p]; x; x = e[x].next) 53 if (dis[p] + e[x].v < dis[(y = e[x].to)]) { 54 dis[y] = dis[p] + e[x].v; 55 if (!v[y]) 56 v[y] = 1, q[(++r) %= N] = y; 57 } 58 v[p] = 0; 59 } 60 } 61 62 int main() { 63 int i, X, Y, Z; 64 m = read(), n = read(), s = read(); 65 T1 = read(), T2 = read(); 66 for (i = 1; i <= m; ++i){ 67 X = read(), Y = read(), Z = read(); 68 Add_Edges(X, Y, Z); 69 } 70 spfa(T1); 71 ans = dis[s] + dis[T2]; 72 spfa(T2); 73 printf("%d ", min(ans, dis[s] + dis[T1])); 74 return 0; 75 }
于是，TLE！！！竟然卡spfa！！！我去！！！
改成Dijkstra就好了。
1 ************************************************************** 2 Problem: 2100 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:228 ms 7 Memory:6796 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cctype> 12 #include <cstring> 13 #include <algorithm> 14 #include <queue> 15 16 using namespace std; 17 const int N = 100005; 18 const int M = 400005; 19 20 struct edges { 21 int next, to, v; 22 edges() {} 23 edges(int _next, int _to, int _v) : next(_next), to(_to), v(_v) {} 24 }e[M]; 25 26 struct heap_node { 27 int v, to; 28 heap_node() {} 29 heap_node(int _v, int _to) : v(_v), to(_to) {} 30 }; 31 inline bool operator < (const heap_node &a, const heap_node &b) { 32 return a.v > b.v; 33 } 34 35 priority_queue <heap_node> h; 36 37 int n, m, s, T1, T2; 38 int first[N], tot; 39 int q[N], l, r, dis[N]; 40 bool v[N]; 41 int ans; 42 43 inline int read() { 44 int x = 0; 45 char ch = getchar(); 46 while (!isdigit(ch)) 47 ch = getchar(); 48 while (isdigit(ch)) { 49 x = x * 10 + ch - '0'; 50 ch = getchar(); 51 } 52 return x; 53 } 54 55 inline void Add_Edges(int x, int y, int z) { 56 e[++tot] = edges(first[x], y, z), first[x] = tot; 57 e[++tot] = edges(first[y], x, z), first[y] = tot; 58 } 59 60 inline void add_to_heap(const int p) { 61 for (int x = first[p]; x; x = e[x].next) 62 if (dis[e[x].to] == -1) 63 h.push(heap_node(e[x].v + dis[p], e[x].to)); 64 } 65 66 void Dijkstra(int S) { 67 memset(dis, -1, sizeof(dis)); 68 while (!h.empty()) h.pop(); 69 dis[S] = 0, add_to_heap(S); 70 int p; 71 while (!h.empty()) { 72 if (dis[h.top().to] != -1) { 73 h.pop(); 74 continue; 75 } 76 p = h.top().to; 77 dis[p] = h.top().v; 78 h.pop(); 79 add_to_heap(p); 80 } 81 } 82 83 int main() { 84 int i, X, Y, Z; 85 m = read(), n = read(), s = read(); 86 T1 = read(), T2 = read(); 87 for (i = 1; i <= m; ++i){ 88 X = read(), Y = read(), Z = read(); 89 Add_Edges(X, Y, Z); 90 } 91 Dijkstra(T1); 92 ans = dis[s] + dis[T2]; 93 Dijkstra(T2); 94 printf("%d ", min(ans, dis[s] + dis[T1])); 95 return 0; 96 }
（p.s. Rank5液！话说hzwer的spfa版是怎么过这题的。。。不明= =）