Travel
The country frog lives in has 1,2,…,n.
Among b minutes to travel.
Find the minimum time to travel from town n.
Input
The input consists of multiple tests. For each test:
The first line contains 1≤ui,vi≤n,ui≠vi).
Output
For each test, write 1 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\n",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\n",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\n",a); 130 else bfsA(); 131 } 132 else 133 { 134 if (b<a) printf("%d\n",b); 135 else bfsB(); 136 } 137 } 138 return 0; 139 }