2017 乌鲁木齐赛区网络赛 J Our Journey of Dalian Ends 费用流-白红宇

2017 乌鲁木齐赛区网络赛 J Our Journey of Dalian Ends 费用流

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发布时间：2019-06-08

本文共 4572 字，大约阅读时间需要 15 分钟。

题目描述：

Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people.

Now our journey of Dalian ends. To be carefully considered are the following questions.

Next month in Xian, an essential lesson which we must be present had been scheduled.

But before the lesson, we need to attend a wedding in Shanghai.

We are not willing to pass through a city twice.

All available expressways between cities are known.

What we require is the shortest path, from Dalian to Xian, passing through Shanghai.

Here we go.

Input Format

There are several test cases.

The first line of input contains an integer

For each test case, the first line contains an integer $m~(m\le 10000)m (m≤10000) which is the number of known expressways.$

Each of the following

The expressway connects two given cities and it is bidirectional.

Output Format

For eact test case, output the shortest path from Dalian to Xian, passing through Shanghai, or output

跑一遍费用流便得到答案。

1 #include
     
        2 #define rep(i,a,b) for(int i=a;i<=b;i++)  3 using namespace std;  4 const int MAXN=101000;  5 const int maxn=20000;  6 const int INF=~0U>>1;  7 int maxflow=0,cost=0;  8 int pre[maxn],vis[maxn],dis[maxn];  9 int S,T,S1; 10 int n,m; 11 int tot=0; 12 int pointer[maxn]; 13 map
      
        mp; 14 int shanghai; 15 struct Edge 16 { 17     int to,next,cap,f,w; 18     Edge() {}; 19     Edge(int b,int c,int nxt,int flow,int weight) {to=b,cap=c,next=nxt,f=flow,w=weight;} 20 }edge[MAXN]; 21 inline void addedge(int a,int b,int c,int w1) 22 { 23     edge[tot]=Edge(b,c,pointer[a],0,w1); 24     pointer[a]=tot++; 25     edge[tot]=Edge(a,0,pointer[b],0,-w1); 26     pointer[b]=tot++; 27 } 28 bool spfa(int s,int t) 29 { 30     queue
       
         q; 31     rep(i,S,T) 32     { 33         dis[i]=INF; 34         vis[i]=false; 35         pre[i]=-1; 36     } 37     dis[S]=0; 38     vis[S]=1; 39     q.push(S); 40     while(!q.empty()) 41     { 42         int u=q.front();q.pop(); 43         vis[u]=0; 44         for(int j=pointer[u];j!=-1;j=edge[j].next) 45         { 46             int v=edge[j].to; 47       //      if(u==92) printf("u=92 v=%d\n",v); 48             if(edge[j].cap-edge[j].f>0&&dis[v]>dis[u]+edge[j].w) 49             { 50                 dis[v]=dis[u]+edge[j].w; 51                 pre[v]=j; 52                 if(!vis[v]) 53                 { 54                     vis[v]=1;q.push(v); 55                 } 56             } 57         } 58     } 59     if(pre[T]==-1) return false; 60     else return true; 61 } 62 int mcmf() 63 { 64     int flow=0; 65     cost=0; 66     while(spfa(S,T)) 67     { 68         int mi=INF; 69         for(int i=pre[T];i!=-1;i=pre[edge[i^1].to]) 70         { 71             mi=min(mi,edge[i].cap-edge[i].f); 72         } 73         for(int i=pre[T];i!=-1;i=pre[edge[i^1].to]) 74         { 75             edge[i].f+=mi; 76             edge[i^1].f-=mi; 77            // printf("edge[%d].w=%d edge[i].to=%d\n",i,edge[i].w,edge[i].to); 78             cost+=edge[i].w*mi; 79         } 80         flow+=mi; 81     } 82     return flow; 83 } 84 void Input() 85 { 86     mp.clear(); 87     scanf("%d",&m); 88     string a; 89     string b; 90     int len; 91     int cnt=1; 92     memset(pointer,-1,sizeof(pointer)); 93     tot=0; 94     rep(i,1,m) 95     { 96         cin>>a>>b>>len; 97         if(mp[a]==0) 98         { 99             mp[a]=cnt;100             addedge(cnt,cnt+1,1,0);101             cnt+=2;102 103         }104         if(mp[b]==0)105         {106             mp[b]=cnt;107             addedge(cnt,cnt+1,1,0);108             cnt+=2;109         }110         addedge(mp[a]+1,mp[b],INF,len);111         addedge(mp[b]+1,mp[a],INF,len);112     }113     S=0;T=cnt;114     addedge(mp["Shanghai"],mp["Shanghai"]+1,1,0);115     addedge(S,mp["Shanghai"],2,0);116     addedge(mp["Xian"]+1,T,1,0);117     addedge(mp["Dalian"]+1,T,1,0);118 }119 int main()120 {121     freopen("in.txt","r",stdin);122     int T;123     scanf("%d",&T);124     rep(t1,1,T)125     {126         Input();127         maxflow=mcmf();128         if(maxflow==2) printf("%d\n",cost);129         else printf("-1\n");130     }131     return 0;132 }