据说标题长可以吸引人们的注意←_←
大家都用spaf。。。不怕被卡吗?
改进的堆优化Dijkstra新鲜出炉了!!!这个板子终于改的既好看又实用了。
1 /************************************************************** 2 Problem: 1681 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:8 ms 7 Memory:872 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 #include <queue> 14 15 using namespace std; 16 struct edges{ 17 int next, to, v; 18 } e[2500]; 19 20 struct heap_node{ 21 int v, to; 22 } NODE; 23 inline bool operator < (const heap_node &a, const heap_node &b){ 24 return a.v > b.v; 25 } 26 27 priority_queue <heap_node> h; 28 int X, Y, Z, F, P, C, time; 29 int first[505], tot, d[505]; 30 int ans[105], cnt; 31 32 inline int read(){ 33 int x = 0, sgn = 1; 34 char ch = getchar(); 35 while (ch < '0' || ch > '9'){ 36 if (ch == '-') sgn = -1; 37 ch = getchar(); 38 } 39 while (ch >= '0' && ch <= '9'){ 40 x = x * 10 + ch - '0'; 41 ch = getchar(); 42 } 43 return sgn * x; 44 } 45 46 inline void add_edge(int X, int Y, int Z){ 47 e[++tot].next = first[X], first[X] = tot; 48 e[tot].to = Y, e[tot].v = Z; 49 } 50 51 void add_Edges(int x, int y, int z){ 52 add_edge(x, y, z); 53 add_edge(y, x, z); 54 } 55 56 inline void add_to_heap(const int p){ 57 for (int x = first[p]; x; x = e[x].next) 58 if (d[e[x].to] == -1){ 59 NODE.v = e[x].v + d[p], NODE.to = e[x].to; 60 h.push(NODE); 61 } 62 } 63 64 void Dijkstra(int S){ 65 memset(d, -1, sizeof(d)); 66 while (!h.empty()) h.pop(); 67 d[S] = 0, add_to_heap(S); 68 int p; 69 while (!h.empty()){ 70 if (d[h.top().to] != -1){ 71 h.pop(); 72 continue; 73 } 74 p = h.top().to; 75 d[p] = h.top().v; 76 h.pop(); 77 add_to_heap(p); 78 } 79 } 80 81 int main(){ 82 F = read(), P = read(), C = read(), time = read(); 83 for (int i = 1; i <= P; ++i){ 84 X = read(), Y = read(), Z = read(); 85 add_Edges(X, Y, Z); 86 } 87 88 Dijkstra(1); 89 for (int i = 1; i <= C; ++i){ 90 X = read(); 91 if (d[X] != -1 && d[X] <= time) 92 ans[++cnt] = i; 93 } 94 printf("%d\n", cnt); 95 for (int i = 1; i <= cnt; ++i) 96 printf("%d\n", ans[i]); 97 return 0; 98 }
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