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：
1 #include <set> 2 #include <map> 3 #include <queue> 4 #include <stack> 5 #include <cmath> 6 #include <bitset> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 typedef long long ll; 17 typedef unsigned long long ull; 18 19 #define lson i<<1 20 #define rson i<<1|1 21 #define bug printf("********* "); 22 #define FIN freopen("D://code//in.txt", "r", stdin); 23 #define debug(x) cout<<"["<<x<<"]" <<endl; 24 #define IO ios::sync_with_stdio(false),cin.tie(0); 25 26 const double eps = 1e-8; 27 const int mod = 10007; 28 const int maxn = 1e5 + 7; 29 const double pi = acos(-1); 30 const int inf = 0x3f3f3f3f; 31 const ll INF = 0x3f3f3f3f3f3f3f; 32 33 inline int read() {//读入挂 34 int ret = 0, c, f = 1; 35 for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar()); 36 if(c == '-') f = -1, c = getchar(); 37 for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0'; 38 if(f < 0) ret = -ret; 39 return ret; 40 } 41 42 int n, m, s, t, u, v, w, k, tot; 43 int d[maxn<<1], cnt[maxn<<1], vis[maxn<<1]; 44 int head[maxn<<1], head1[maxn<<1]; 45 pair<int, int> P; 46 47 struct b { 48 int v, w, next; 49 }biao[maxn<<1]; 50 51 struct node { 52 int g, h; 53 int to; 54 bool operator < (node a) const { 55 return (a.h + a.g) < h + g; 56 } 57 }; 58 59 void add(int u, int v, int w) { 60 biao[tot].v = v; 61 biao[tot].w = w; 62 biao[tot].next = head[u]; 63 head[u] = tot++; 64 65 biao[tot].v = u; 66 biao[tot].w = w; 67 biao[tot].next = head1[v]; 68 head1[v] = tot++; 69 } 70 71 void init() { 72 tot = 0; 73 memset(head, -1, sizeof(head)); 74 memset(head1, -1, sizeof(head1)); 75 } 76 77 void spfa() { 78 memset(vis, 0, sizeof(vis)); 79 memset(d, inf, sizeof(d)); 80 d[t] = 0; 81 vis[t] = 1; 82 queue<int> q; 83 q.push(t); 84 while(!q.empty()) { 85 int u = q.front(); 86 q.pop(); 87 vis[u] = 0; 88 for(int i = head1[u]; i != -1; i = biao[i].next) { 89 int v = biao[i].v; 90 if(d[v] > d[u] + biao[i].w) { 91 d[v] = d[u] + biao[i].w; 92 if(!vis[v]) { 93 q.push(v); 94 vis[v] = 1; 95 } 96 } 97 } 98 } 99 } 100 101 int AA() { 102 memset(cnt, 0, sizeof(cnt)); 103 priority_queue<node> Q; 104 node p, q; 105 p.g = 0; 106 p.h = d[s]; 107 p.to = s; 108 Q.push(p); 109 while(!Q.empty()) { 110 p = Q.top(), Q.pop(); 111 cnt[p.to]++; 112 if(cnt[p.to] > k) continue; 113 if(cnt[t] == k) return p.g; 114 int u = p.to; 115 for(int i = head[u]; i != -1; i = biao[i].next) { 116 int v = biao[i].v; 117 q.to = v; 118 q.g = p.g + biao[i].w; 119 q.h = d[v]; 120 Q.push(q); 121 } 122 } 123 return -1; 124 } 125 126 int main() { 127 //FIN; 128 scanf("%d%d", &n, &m); 129 init(); 130 while(m--) { 131 scanf("%d%d%d", &u, &v, &w); 132 add(u, v, w); 133 } 134 scanf("%d%d%d", &s, &t, &k); 135 spfa(); 136 if(s == t) k++; 137 int ans = AA(); 138 printf("%d ", ans); 139 return 0; 140 }