http://acm.hdu.edu.cn/showproblem.php?pid=5521
Meeting
Problem Description
Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated his
fences they were separated into different blocks. John's farm are divided into n blocks labelled from 1 to n.
Bessie lives in the first block while Elsie lives in the n-th one. They have a map of the farm
which shows that it takes they ti minutes to travel from a block in Ei to another block
in Ei where Ei (1≤i≤m) is a set of blocks. They want to know how soon they can meet each other
and which block should be chosen to have the meeting.
fences they were separated into different blocks. John's farm are divided into n blocks labelled from 1 to n.
Bessie lives in the first block while Elsie lives in the n-th one. They have a map of the farm
which shows that it takes they ti minutes to travel from a block in Ei to another block
in Ei where Ei (1≤i≤m) is a set of blocks. They want to know how soon they can meet each other
and which block should be chosen to have the meeting.
Input
The first line contains an integer T (1≤T≤6), the number of test cases. Then T test cases
follow.
The first line of input contains n and m. 2≤n≤105. The following m lines describe the sets Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109) and Si (Si>0) firstly. Then Si integer follows which are the labels of blocks in Ei. It is guaranteed that ∑mi=1Si≤106.
Output
For each test case, if they cannot have the meeting, then output "Evil John" (without quotes) in one line.
Otherwise, output two lines. The first line contains an integer, the time it takes for they to meet.
The second line contains the numbers of blocks where they meet. If there are multiple
optional blocks, output all of them in ascending order.
The second line contains the numbers of blocks where they meet. If there are multiple
optional blocks, output all of them in ascending order.
Sample Input
2
5 4
1 3 1 2 3
2 2 3 4
10 2 1 5
3 3 3 4 5
3 1
1 2 1 2
Sample Output
Case #1: 3
3 4
Case #2: Evil John
Hint
题意：给出n个点还有m个集合，每个集合给出一个dis和s个点，这个集合里面的点去其他点的花费是dis，然后求从1到n和从n到1同时走到相同的点的最短距离。
思路：由于给出的点的关系是集合，而且数据范围不允许我们两两建边，看了一下别人的结题报告，发现一个比较巧妙的方法：我们可以在这些点之外再弄出m个点来，每个集合的点和n之外的一个点存在边，这样的话每个集合的建边就是2*s而已，这样的建边让距离乘以2了，我们可以先保留这个距离，到输出答案时候再除以2.
1 #include <cstdio> 2 #include <algorithm> 3 #include <queue> 4 #include <cstring> 5 #include <iostream> 6 using namespace std; 7 typedef long long LL; 8 #define N 1000005 9 #define M 2000005 10 const LL inf = 1e17; 11 /* 12 最短路 13 */ 14 struct node 15 { 16 int v, nxt; 17 LL w; 18 }edge[M]; 19 20 struct nd 21 { 22 int p; 23 LL d; 24 nd(){} 25 nd(LL _d, int _to):d(_d), p(_to) {} 26 bool operator < (const nd &a) const { 27 return d > a.d; 28 } 29 }; 30 31 int tot, n, m, nn, used[N], head[N]; 32 LL da[N], db[N]; 33 34 void add(int u, int v, LL w) 35 { 36 edge[tot].v = v; 37 edge[tot].w = w; 38 edge[tot].nxt = head[u]; 39 head[u] = tot++; 40 } 41 42 void Dijkstra(int s, LL d[]) 43 { 44 priority_queue<nd> que; 45 while(!que.empty()) que.pop(); 46 for(int i = 0 ; i <= nn; i++) d[i] = inf; 47 memset(used, 0, sizeof(used)); 48 d[s] = 0; 49 que.push(nd(d[s], s)); 50 while(!que.empty()) { 51 nd top = que.top(); que.pop(); 52 int u = top.p; 53 if(used[u]) continue; 54 used[u] = 1; 55 for(int k = head[u]; ~k; k = edge[k].nxt) { 56 int v = edge[k].v; 57 int w = edge[k].w; 58 if(d[u]+w < d[v]) { 59 d[v] = d[u] + w; 60 que.push(nd(d[v], v)); 61 } 62 } 63 } 64 } 65 66 int main() 67 { 68 int t; 69 scanf("%d", &t); 70 for(int cas = 1; cas <= t; cas++) { 71 scanf("%d%d", &n, &m); 72 memset(head, -1, sizeof(head)); 73 tot = 0; 74 nn = n; 75 for(int i = 0; i < m; i++) { 76 LL dis; 77 int num, a; 78 //就是这里了，在外面建一个点，然后集合里面所有点都和它相连 79 nn++; 80 scanf("%I64d%d", &dis, &num); 81 for(int j = 0; j < num; j++) { 82 scanf("%d", &a); 83 add(a, nn, dis); 84 add(nn, a, dis); 85 } 86 } 87 88 Dijkstra(1, da); 89 Dijkstra(n, db); 90 LL ans = inf; 91 92 93 for(int i = 1; i <= n; i++) 94 ans = min(ans, max(da[i], db[i])); 95 96 printf("Case #%d: ", cas); 97 if(ans==inf) printf("Evil John "); 98 else{ 99 //因为求出的距离是两倍，所以要除以二 100 printf("%I64d ", ans/2); 101 bool f = 0; 102 for(int i = 1; i <= n; i++) { 103 if(max(da[i], db[i]) == ans){ 104 if(f) printf(" "); 105 f = 1; 106 printf("%d", i); 107 } 108 } 109 puts(""); 110 } 111 } 112 return 0; 113 }