Cow Relays
- Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
- Total Submission(s): 80 Accepted Submission(s): 13
Farmer John has formed a relay team for a race by choosing K (2 ≤ K ≤ 40) of his cows. The race is run on FJ's farm which has N (4 ≤ N < 800) fields numbered 1..N and M (1 ≤ M ≤ 4000) unique bidirectional pathways that connect pairs of different fields. You will be given the time it takes a cow to traverse each pathway.
The first cow begins the race in field #1 and runs to the finish line in field #N. As soon as the first cow finishes, the next cow then starts from field #1 and runs to field #N and so on. For this race, no two cows can follow precisely the same route (a route is a sequence of fields).
Write a program which computes the minimum possible time required for FJ's relay team. It is guaranteed that some minimum possible time exists. Any cows can revisit a path in her trip to the other barn if that turns out to be required for a "best" solution. As soon as a cow enters field #N, her relay leg is finished.
* Line 1: One line with three integers: K, N, and M
* Lines 2..M+1: Each line contains three integers describing a path: the starting field, the ending field, and the integer time to traverse the path (in the range 1..9500).
One line with a single integer that is the minimum possible time to run a relay.
4 5 8
1 2 1
1 3 2
1 4 2
2 3 2
2 5 3
3 4 3
3 5 4
4 5 6
23
Namely: Cow 1: 1->2->5 4
Cow 2: 1->3->5 6
Cow 3: 1->2->1->2->5 6
Cow 4: 1->2->3->5 7
题意概括：
给出一个具有 N 个顶点 M 条边的无向图，求从起点 1 到 终点 N 的K条最短路径长度之和；
解题思路：
BFS求最短路，但是顶点可以重复入队，但只允许入队 K 次，因为入队几次就说明了是求到了当前第K的最短路，我们只需要前K个，所以后面的不再入队。
优先队列能保证求到的K条路径一定是最优的。
注意：标记入队次数是在处理该节点时才标记，而不是该节点入队时标记，因为入队不一定被处理，只有被处理了才算用到了该节点来求最短路径。
AC code：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 #include <vector> 7 #define INF 0x3f3f3f3f 8 #define LL long long 9 using namespace std; 10 int N, M, K, ans; 11 const int MAXN = 900; 12 const int MAXM = 8005; 13 int in_c[MAXN]; 14 15 struct Edge 16 { 17 int v, nxt, w; 18 }edge[MAXM]; 19 int head[MAXN], cnt; 20 21 struct data 22 { 23 int x, len; 24 bool operator<(const data &a) const { 25 return len > a.len; ///最小值优先 26 } 27 }; 28 priority_queue<struct data>que; 29 30 31 void init(){ 32 memset(head, -1, sizeof(head)); 33 memset(in_c, 0, sizeof(in_c)); 34 cnt = 0; 35 ans = 0; 36 } 37 38 void add(int from, int to, int cost) 39 { 40 edge[cnt].v = to; 41 edge[cnt].w = cost; 42 edge[cnt].nxt = head[from]; 43 head[from] = cnt++; 44 } 45 46 void BFS() 47 { 48 //while(!que.empty()) que.pop(); 49 struct data it, temp; 50 it.x = 1; 51 it.len = 0; 52 //in_c[1]++; 53 que.push(it); 54 int sum = 0; 55 56 while(!que.empty()){ 57 it = que.top();que.pop(); 58 if(it.x == N){ 59 sum++; 60 ans+=it.len; 61 if(sum == K) return; 62 continue; 63 } 64 if(++in_c[it.x] > K) continue; 65 66 for(int i = head[it.x]; i != -1; i = edge[i].nxt){ 67 int to = edge[i].v; 68 //printf("to:%d ", to); 69 //if(in_c[to] < K){ 70 //in_c[to]++; 71 temp.x = to; 72 temp.len = it.len+edge[i].w; 73 que.push(temp); 74 //} 75 } 76 } 77 78 } 79 80 int main() 81 { 82 int u, v, w; 83 //while(~scanf("%d %d %d", &K, &N, &M)){ 84 scanf("%d %d %d", &K, &N, &M); 85 init(); 86 for(int i = 1; i <= M; i++){ 87 scanf("%d %d %d", &u, &v, &w); 88 add(u, v, w); 89 add(v, u, w); 90 } 91 92 BFS(); 93 94 printf("%d ", ans); 95 //} 96 97 return 0; 98 }