Travel
The country frog lives in has nn towns which are conveniently numbered by 1,2,…,n1,2,…,n.
Among n(n−1)2n(n−1)2 pairs of towns, mm of them are connected by bidirectional highway, which needs aa minutes to travel. The other pairs are connected by railway, which needs bb minutes to travel.
Find the minimum time to travel from town 11 to town nn.
Input
The input consists of multiple tests. For each test:
The first line contains 44 integers n,m,a,bn,m,a,b (2≤n≤105,0≤m≤5⋅105,1≤a,b≤1092≤n≤105,0≤m≤5⋅105,1≤a,b≤109). Each of the following mm lines contains 22integers ui,viui,vi, which denotes cities uiui and vivi are connected by highway. (1≤ui,vi≤n,ui≠vi1≤ui,vi≤n,ui≠vi).
Output
For each test, write 11 integer which denotes the minimum time.
Sample Input
3 2 1 3
1 2
2 3
3 2
2 3 1 2
2 3
Sample Output
2
3
//题意: n , m, a, b ，其中 n 代表点数，并且是个完全图，m 是权值为 a 的边数，其余的边权值为 b ，问 1--n 的最短路
题解:如果 1 和 n 之间连边为 a 那么答案一定为 a 与一条最短的全由b组成的路径的较小者，如果 1 和 n 之间连边为b，那么答案一定
为b和一条最短的全由a组成的路径的较小者。对于第1种情况直接bfs就可以，第二种情况由于边数较多，不能直接bfs
从1开始搜索与其相连的边权为b的边，用set维护一下，由于每个点只入队1次，复杂度算是 nlogn ，叉姐的题很有意思
300ms
1 # include <cstdio> 2 # include <cstring> 3 # include <cstdlib> 4 # include <iostream> 5 # include <vector> 6 # include <queue> 7 # include <stack> 8 # include <map> 9 # include <bitset> 10 # include <sstream> 11 # include <set> 12 # include <cmath> 13 # include <algorithm> 14 # pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 # define LL long long 17 # define pr pair 18 # define mkp make_pair 19 # define lowbit(x) ((x)&(-x)) 20 # define PI acos(-1.0) 21 # define INF 0x3f3f3f3f3f3f3f3f 22 # define eps 1e-8 23 # define MOD 1000000007 24 25 inline int scan() { 26 int x=0,f=1; char ch=getchar(); 27 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 28 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 29 return x*f; 30 } 31 inline void Out(int a) { 32 if(a<0) {putchar('-'); a=-a;} 33 if(a>=10) Out(a/10); 34 putchar(a%10+'0'); 35 } 36 # define MX 100005 37 /**************************/ 38 struct Edge 39 { 40 int v,nex; 41 }edge[MX*10]; 42 43 int n,m,a,b,ip; 44 int hlist[MX]; 45 LL dis[MX]; 46 bool vis[MX]; 47 void addedge(int u,int v) 48 { 49 edge[ip]= (Edge){v,hlist[u]}; 50 hlist[u]=ip++; 51 edge[ip]= (Edge){u,hlist[v]}; 52 hlist[v]=ip++; 53 } 54 55 void bfsB() // 1-n 连b边 56 { 57 dis[n]=INF; 58 memset(vis,0,sizeof(vis)); 59 queue<int> Q; 60 Q.push(1); 61 dis[1]=0; 62 vis[1]=1; 63 while (!Q.empty()) 64 { 65 int u = Q.front(); Q.pop(); 66 for (int i=hlist[u];i!=-1;i=edge[i].nex) 67 { 68 int v = edge[i].v; 69 if (!vis[v]) 70 { 71 dis[v]=dis[u]+1; 72 Q.push(v); 73 vis[v]=1; 74 } 75 } 76 if (dis[n]!=INF) break; 77 } 78 printf("%lld ",min(dis[n]*a,(LL)b)); 79 } 80 81 void bfsA() //1-n 连 a 边 82 { 83 dis[n]=INF; 84 set<int> st,ts; 85 for (int i=2;i<=n;i++) st.insert(i); 86 set<int>::iterator it; 87 queue<int> Q; 88 Q.push(1); 89 dis[1]=0; 90 while (!Q.empty()) 91 { 92 int u = Q.front(); Q.pop(); 93 for (int i=hlist[u];i!=-1;i=edge[i].nex) 94 { 95 int v=edge[i].v; 96 if (st.count(v)==0) continue; 97 st.erase(v); ts.insert(v); 98 } 99 for (it=st.begin();it!=st.end();it++) 100 { 101 dis[*it] = dis[u]+1; 102 Q.push(*it); 103 } 104 if (dis[n]!=INF) break; 105 st.swap(ts); 106 ts.clear(); 107 } 108 printf("%lld ",min(dis[n]*b,(LL)a)); 109 } 110 111 112 int main() 113 { 114 while(scanf("%d%d%d%d",&n,&m,&a,&b)!=EOF) 115 { 116 memset(hlist,-1,sizeof(hlist)); 117 ip=0; 118 bool flag=0; 119 for (int i=0;i<m;i++) 120 { 121 int u = scan(); 122 int v = scan(); 123 addedge(u,v); 124 if (u>v) swap(u,v); 125 if (u==1&&v==n) flag=1; 126 } 127 if (flag) 128 { 129 if (a<b) printf("%d ",a); 130 else bfsA(); 131 } 132 else 133 { 134 if (b<a) printf("%d ",b); 135 else bfsB(); 136 } 137 } 138 return 0; 139 }