Remmarguts' Date（POJ2449+最短路+A*算法）

题目链接：http://poj.org/problem?id=2449

题目：

题意：求有向图两点间的k短路。

思路：最短路+A*算法

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 1e5 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 inline int read() {//读入挂
     int ret = , c, f = ;
     for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
     if(c == '-') f = -, c = getchar();
     for(; isdigit(c); c = getchar()) ret = ret *  + c - '';
     if(f < ) ret = -ret;
     return ret;
 }

 int n, m, s, t, u, v, w, k, tot;
 int d[maxn<<], cnt[maxn<<], vis[maxn<<];
 int head[maxn<<], head1[maxn<<];
 pair<int, int> P;

 struct b {
     int v, w, next;
 }biao[maxn<<];

 struct node {
     int g, h;
     int to;
     bool operator < (node a) const {
         return (a.h + a.g) < h + g;
     }
 };

 void add(int u, int v, int w) {
     biao[tot].v = v;
     biao[tot].w = w;
     biao[tot].next = head[u];
     head[u] = tot++;

     biao[tot].v = u;
     biao[tot].w = w;
     biao[tot].next = head1[v];
     head1[v] = tot++;
 }

 void init() {
     tot = ;
     memset(head, -, sizeof(head));
     memset(head1, -, sizeof(head1));
 }

 void spfa() {
     memset(vis, , sizeof(vis));
     memset(d, inf, sizeof(d));
     d[t] = ;
     vis[t] = ;
     queue<int> q;
     q.push(t);
     while(!q.empty()) {
         int u = q.front();
         q.pop();
         vis[u] = ;
         for(int i = head1[u]; i != -; i = biao[i].next) {
             int v = biao[i].v;
             if(d[v] > d[u] + biao[i].w) {
                 d[v] = d[u] + biao[i].w;
                 if(!vis[v]) {
                     q.push(v);
                     vis[v] = ;
                 }
             }
         }
     }
 }

 int AA() {
     memset(cnt, , sizeof(cnt));
     priority_queue<node> Q;
     node p, q;
     p.g = ;
     p.h = d[s];
     p.to = s;
     Q.push(p);
     while(!Q.empty()) {
         p = Q.top(), Q.pop();
         cnt[p.to]++;
         if(cnt[p.to] > k) continue;
         if(cnt[t] == k) return p.g;
         int u = p.to;
         for(int i = head[u]; i != -; i = biao[i].next) {
             int v = biao[i].v;
             q.to = v;
             q.g = p.g + biao[i].w;
             q.h = d[v];
             Q.push(q);
         }
     }
     return -;
 }

 int main() {
     //FIN;
     scanf("%d%d", &n, &m);
     init();
     while(m--) {
         scanf("%d%d%d", &u, &v, &w);
         add(u, v, w);
     }
     scanf("%d%d%d", &s, &t, &k);
     spfa();
     if(s == t) k++;
     int ans = AA();
     printf("%d\n", ans);
     return ;
 }