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

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。

/**************************************************************
    Problem: 1681
    User: rausen
    Language: C++
    Result: Accepted
    Time:8 ms
    Memory:872 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
 
using namespace std;
struct edges{
    int next, to, v;
} e[2500];
 
struct heap_node{
    int v, to;
} NODE;
inline bool operator < (const heap_node &a, const heap_node &b){
    return a.v > b.v;
}
 
priority_queue <heap_node> h;
int X, Y, Z, F, P, C, time;
int first[505], tot, d[505];
int ans[105], cnt;
 
inline int read(){
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
inline void add_edge(int X, int Y, int Z){
    e[++tot].next = first[X], first[X] = tot;
    e[tot].to = Y, e[tot].v = Z;
}
 
void add_Edges(int x, int y, int z){
    add_edge(x, y, z);
    add_edge(y, x, z);
}
 
inline void add_to_heap(const int p){
    for (int x = first[p]; x; x = e[x].next)
        if (d[e[x].to] == -1){
            NODE.v = e[x].v + d[p], NODE.to = e[x].to;
            h.push(NODE);
        }
}
 
void Dijkstra(int S){
    memset(d, -1, sizeof(d));
    while (!h.empty()) h.pop();
    d[S] = 0, add_to_heap(S);
    int p;
    while (!h.empty()){
        if (d[h.top().to] != -1){
            h.pop();
            continue;
        }
        p = h.top().to;
        d[p] = h.top().v;
        h.pop();
        add_to_heap(p);
    }
}
 
int main(){
    F = read(), P = read(), C = read(), time = read();
    for (int i = 1; i <= P; ++i){
        X = read(), Y = read(), Z = read();
        add_Edges(X, Y, Z);
    }
     
    Dijkstra(1);
    for (int i = 1; i <= C; ++i){
        X = read();
        if (d[X] != -1 && d[X] <= time)
            ans[++cnt] = i;
    }
    printf("%d
", cnt);
    for (int i = 1; i <= cnt; ++i)
        printf("%d
", ans[i]);
    return 0;
}