BZOJ1681 [Usaco2005 Mar]Checking an Alibi 不在场的证明

 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 }