Meeting 加虛擬邊
fences they were separated into different blocks. John‘s farm are divided into nn blocks labelled from 11 to nn.
Bessie lives in the first block while Elsie lives in the nn-th one. They have a map of the farm
which shows that it takes they
in EiEi where Ei (1≤i≤m)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.
InputThe first line contains an integer T (1≤T≤6)T (1≤T≤6), the number of test cases. Then
follow.
The first line of input contains nn and mm. 2≤n≤1052≤n≤105. The following mm lines describe the sets Ei (1≤i≤m)Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109)ti(1≤ti≤109) and Si (Si>0)Si (Si>0) firstly. Then SiSi integer follows which are the labels of blocks in Ei
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.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
可以將給頂集合的元素連到一個虛擬結點上,求出最短路來再/2,這樣避免了大量的重復加邊,還避免了小數
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<sstream> #include<algorithm> #include<queue> #include<vector> #include<cmath> #include<map> #include<stack> #include<set> #include<memory> #include<bitset> #include<string> #include<functional> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int MAXN = 5e5 ; #define INF 0x3f3f3f3f /* 連接虛擬結點 到該點的距離為L 求出最短路/2 避免小數! */ LL T, d, n, m, cnt; struct edge { edge(LL _a,LL _b):to(_a),cost(_b){} LL to, cost; }; vector<edge>E[MAXN]; LL dist1[MAXN], dist2[MAXN]; bool vis[MAXN]; void addedge(LL f,LL to,LL dis) { E[f].push_back(edge(to, dis)); E[to].push_back(edge(f, dis)); } void init() { for (LL i = 0; i < MAXN; i++) E[i].clear(); } void spfa(LL beg, LL lowcost[]) { queue<LL> q; memset(vis, false, sizeof(vis)); for (int i = 0; i <= n + m; i++) lowcost[i] = INF; lowcost[beg] = 0; vis[beg] = true; q.push(beg); while (!q.empty()) { LL f = q.front(); q.pop(); vis[f] = false; for (int i = 0; i < E[f].size(); i++) { if (lowcost[E[f][i].to] > lowcost[f] + E[f][i].cost) { lowcost[E[f][i].to] = lowcost[f] + E[f][i].cost; if (!vis[E[f][i].to]) { vis[E[f][i].to] = true; q.push(E[f][i].to); } } } } } int main() { scanf("%lld", &T); for(LL cas = 1;cas <= T; cas++) { init(); scanf("%lld%lld", &n, &m); LL tmp, tt; for (LL i = 1; i <= m; i++) { scanf("%lld%lld", &d, &tmp); while (tmp--) { scanf("%lld", &tt); addedge(tt, n + i, d); } } spfa(1, dist1); spfa(n , dist2); LL ans = INF; for (int i = 0; i <= n; i++) ans = min(ans, max(dist1[i], dist2[i])); if (ans == INF) printf("Case #%lld: Evil John\n", cas); else { printf("Case #%lld: %lld\n", cas, ans / 2); bool f = false; for (int i = 1; i <= n; i++) { if (max(dist1[i], dist2[i]) == ans) { if (!f) printf("%d", i), f = true; else printf(" %d", i); } } printf("\n"); } } }
Meeting 加虛擬邊