POJ3013-Big Christmas Tree-最短路

Posted 2020-06-18 Helica

题意：给出一个图，每个节点都有权值，每条边也有费用。要求建立一颗树，使总花费最小。树上每连一条边的花费定义为孩子节点权值和×此边费用。

做法：分析可知，最终的答案为所有节点的权值×到根节点的距离。可以知道当距离最短时，花费最小。

　　　 于是用Dijkstra+优先队列优化就可以搞定了。这题有些卡时间。最后还要注意使用long long，特判n=0和n=1。

  1 /*--------------------------------------------------------------------------------------*/
  2 //        Helica‘s header 
  3 //        Second Edition
  4 //        2015.11.7
  5 //
  6 #include <algorithm>
  7 #include <iostream>
  8 #include <cstring>
  9 #include <ctype.h>
 10 #include <cstdlib>
 11 #include <cstdio>
 12 #include <vector>
 13 #include <string>
 14 #include <queue>
 15 #include <stack>
 16 #include <cmath>
 17 #include <set>
 18 #include <map>
 19 
 20 //debug function for a N*M array 
 21 #define debug_map(N,M,G) printf("\n");for(int i=0;i<(N);i++) 22 {for(int j=0;j<(M);j++){ 23 printf("%d",G[i][j]);}printf("\n");}                
 24 //debug function for int,float,double,etc.
 25 #define debug_var(X) cout<<#X"="<<X<<endl;
 26 /*--------------------------------------------------------------------------------------*/
 27 using namespace std;
 28 
 29 const int maxn = 5e4+10;
 30 const long long INF = 0x3f3f3f3f3f3f3f3f;
 31 int N,M,T,S;
 32 
 33 struct qnode
 34 {
 35     int v,c;
 36     qnode(int _v=0,int _c=0):v(_v),c(_c){}
 37     bool operator < (const qnode &r) const
 38     {return c>r.c;}
 39 };
 40 
 41 struct Edge
 42 {
 43     int to,next;
 44     int cost;
 45 }edge[4*maxn];
 46 
 47 int head[maxn],tol,weight[maxn];
 48 long long dis[maxn];
 49 bool vis[maxn];
 50 priority_queue<qnode> que;
 51 
 52 void Dijkstra(int n,int start)
 53 {
 54     memset(vis,false,sizeof vis);
 55     for(int i=1;i<=n;i++) dis[i] = INF;
 56     dis[start] = 0;
 57     while(!que.empty()) que.pop();
 58 
 59     que.push(qnode(start,0));
 60     qnode cur;
 61     while(!que.empty())
 62     {
 63         cur = que.top();
 64         que.pop();
 65         int u = cur.v;
 66         if(vis[u]) continue;
 67         vis[u] = true;
 68 
 69         for(int i=head[u];~i;i=edge[i].next)
 70         {
 71             int v = edge[i].to;
 72             int cost = edge[i].cost;
 73             //printf("u:%d v:%d cost:%d\n",u,v,cost);
 74             if(!vis[v] && dis[v]>dis[u]+cost)
 75             {
 76                 dis[v] = dis[u]+cost;
 77                 que.push(qnode(v,dis[v]));
 78             }
 79         }
 80     }
 81 }
 82 
 83 void add_edge(int u,int v,int cost)
 84 {
 85     edge[tol].to = u;
 86     edge[tol].next = head[v];
 87     edge[tol].cost = cost;
 88     head[v] = tol++;
 89 
 90     edge[tol].to = v;
 91     edge[tol].next = head[u];
 92     edge[tol].cost = cost;
 93     head[u] = tol++;
 94 }
 95 
 96 int main()
 97 {
 98     scanf("%d",&T);
 99     while(T--)
100     {
101         scanf("%d%d",&N,&M);
102         for(int i=1;i<=N;i++) scanf("%d",&weight[i]);
103         memset(head,-1,sizeof head);
104         memset(vis,false,sizeof vis);
105         tol = 0;
106         for(int i=0;i<M;i++)
107         {
108             int a,b,c;
109             scanf("%d%d%d",&a,&b,&c);
110             add_edge(a,b,c);
111         }
112         if(N==0||N==1)
113         {
114             printf("0\n");
115             continue;
116         }
117         Dijkstra(N,1);
118 
119         long long ans = 0;
120         bool flag = true;
121         for(int i=2;i<=N;i++)
122         {
123             if(dis[i] == INF) 
124             {
125                 flag = false;
126                 break;
127             }
128             else
129                 ans += weight[i]*dis[i];
130         }
131 
132         if(!flag) printf("No Answer\n");
133         else printf("%lld\n",ans);
134     }
135 }