Flight
Time Limit : 20000/10000ms (Java/Other) Memory Limit : 65535/65535K (Java/Other)
Total Submission(s) : 5 Accepted Submission(s) : 1
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Problem Description
Recently, Shua Shua had a big quarrel with his GF. He is so upset that he decides to take a trip to some other city to avoid meeting her. He will travel only by air and he can go to any city if there exists a flight and it can help him reduce the total cost to the destination. There's a problem here: Shua Shua has a special credit card which can reduce half the price of a ticket ( i.e. 100 becomes 50, 99 becomes 49. The original and reduced price are both integers. ). But he can only use it once. He has no idea which flight he should choose to use the card to make the total cost least. Can you help him?
Input
There are no more than 10 test cases. Subsequent test cases are separated by a blank line.
The first line of each test case contains two integers N and M ( 2 <= N <= 100,000
0 <= M <= 500,000 ), representing the number of cities and flights. Each of the following M lines contains "X Y D" representing a flight from city X to city Y with ticket price D ( 1 <= D <= 100,000 ). Notice that not all of the cities will appear in the list! The last line contains "S E" representing the start and end city. X, Y, S, E are all strings consisting of at most 10 alphanumeric characters.
The first line of each test case contains two integers N and M ( 2 <= N <= 100,000
0 <= M <= 500,000 ), representing the number of cities and flights. Each of the following M lines contains "X Y D" representing a flight from city X to city Y with ticket price D ( 1 <= D <= 100,000 ). Notice that not all of the cities will appear in the list! The last line contains "S E" representing the start and end city. X, Y, S, E are all strings consisting of at most 10 alphanumeric characters.
Output
One line for each test case the least money Shua Shua have to pay. If it's impossible for him to finish the trip, just output -1.
Sample Input
4 4 Harbin Beijing 500 Harbin Shanghai 1000 Beijing Chengdu 600 Shanghai Chengdu 400 Harbin Chengdu 4 0 Harbin Chengdu
Sample Output
800 -1
Hint
In the first sample, Shua Shua should use the card on the flight from
Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the
least total cost 800. In the second sample, there's no way for him to get to
Chengdu from Harbin, so -1 is needed.
Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the
least total cost 800. In the second sample, there's no way for him to get to
Chengdu from Harbin, so -1 is needed.
/* 这题刚开始我以为只要先找到最短路径，然后再把机票最贵的折半就行了， 后来发现，不一定。。。。这样最便宜(。﹏。) 上网搜了一下，有的是spfa+dp有的分层图，我决定把两种都写一遍(๑•̀ㅂ•́)و✧ */ /* spfa+dp dis[0][i]表示到i点折半的花费 dis[1][i]表示到i点没用过折半的花费 这题还要注意的是花费总额已经超出int范围，要用long long 然而我inf最大值没有改成LLONG_MAX错了好几次 */ #include <iostream> #include<cstdio> #include<queue> #include<map> #include<vector> #include<algorithm> #include<climits> using namespace std; long long dis[100005][2]; bool vis[100005]; struct node { int num,d; node(int a,int b){num=a; d=b;} }; vector<node> s[100005]; int n,m; const long long inf=LLONG_MAX; map<string,int> mp; void spfa(int k) { for(int i=1;i<=n;i++) dis[i][1]=dis[i][0]=inf,vis[i]=0; queue<int> Q; Q.push(k); vis[k]=1; dis[k][1]=dis[k][0]=0; while(!Q.empty()) { int u=Q.front(); Q.pop(); vis[u]=0; for(int i=0;i<s[u].size();i++) { bool flag=0; if (dis[s[u][i].num][1]>dis[u][1]+s[u][i].d) { flag=1; dis[s[u][i].num][1]=dis[u][1]+s[u][i].d; } if (dis[s[u][i].num][0]>dis[u][0]+s[u][i].d || dis[s[u][i].num][0]>dis[u][1]+s[u][i].d/2) { dis[s[u][i].num][0]=min(dis[u][0]+s[u][i].d,dis[u][1]+s[u][i].d/2); flag=1; } if(flag && !vis[s[u][i].num]) { vis[s[u][i].num]=1; Q.push(s[u][i].num); } } } return; } int main() { while(~scanf("%d%d",&n,&m)) { int l=0,d; char ch2[15],ch1[15]; for(int i=1;i<=n;i++) s[i].clear(); mp.clear(); for(int i=1;i<=m;i++) { scanf("%s %s %d",&ch1,&ch2,&d); if (!mp[ch1]) mp[ch1]=++l; if (!mp[ch2]) mp[ch2]=++l; s[mp[ch1]].push_back( node(mp[ch2],d) ); } scanf("%s %s",&ch1,&ch2); if (!mp[ch1]) mp[ch1]=++l;//可能m=0，所以有可能开始和结束城市都没有编号 if (!mp[ch2]) mp[ch2]=++l; int scity=mp[ch1]; //开始的城市编号 int ecity=mp[ch2]; //相要到达的城市编号 spfa(scity); if (dis[ecity][0]==inf) printf("-1\n"); else printf("%lld\n",dis[ecity][0]); } return 0; }
最后还是拉了一段代码，(⊙﹏⊙)b
/* 转自：http://yomean.blog.163.com/blog/static/189420225201110282390985/ 一看就想到了分层图，不过如果用分层图，有点杀鸡用牛刀的感觉，因为只有两层。但我还是写了，最后AC了。不过网上很多人都是用建反两向边求解。 而对于分层图求最短路径问题，我们要注意的是，层与层之间的连线都是单向的，而且是从下一层指向上一层，而我们求最短路径的时候，起点总是在下一层，而终点总是在上一层，所以我们可以将经过层与层之间的特殊边的数目控制在n - 1(n是层数）。 */ #include<iostream> #include<cstdio> #include<string> #include<queue> #include<map> #include<vector> #define N 100005 #define inf (_I64_MAX)/2 using namespace std; int n,m; int head[2*N],vis[2*N]; int now,index,k; __int64 dis[2*N]; char name[N][10]; map<string,int>M; struct node{ int v,w,next; }edge[15*N]; void addedge(int u,int v,int w) { edge[index].v=v; edge[index].w=w; edge[index].next=head[u]; head[u]=index++; } struct cmp{ bool operator()(int a,int b){ return dis[a]>dis[b]; } }; priority_queue<int,vector<int>,cmp>Q; void init() { while(!Q.empty()) Q.pop(); M.erase(M.begin(),M.end()); for(int i=0;i<2*n;i++){ vis[i]=false; head[i]=-1; } now=1; index=0; } void dij(int s,int e) { for(int i=0;i<=2*n;i++){ dis[i]=inf;vis[i]=false; } dis[s]=0; vis[s]=true; Q.push(s); while(!Q.empty()){ int u=Q.top(); Q.pop(); if(u==e){ printf("%I64d\n",dis[u]); return; } for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; int w=edge[i].w; if(!vis[v] && dis[v]>dis[u]+w){ dis[v]=dis[u]+w; Q.push(v); } } } } int main(void) { string a,b; int x,y,c; while(cin>>n>>m) { init(); for(int i=0;i<m;i++){ cin>>a>>b>>c; if(M.find(a)==M.end()) M[a]=now++; if(M.find(b)==M.end()) M[b]=now++; addedge(M[a],M[b],c); addedge(M[a]+n,M[b]+n,c); addedge(M[a]+n,M[b],c/2); } cin>>a>>b; __int64 ans=inf; if(M.find(a)==M.end() || M.find(b)==M.end()){ puts("-1");continue; } else dij(M[a]+n,M[b]); } return 0; }