HDU 4318 Power transmission（最短路）

http://acm.hdu.edu.cn/showproblem.php?pid=4318

题意：

给出运输路线，每条路线运输时都会损失一定百分比的量，给定起点、终点和初始运输量，问最后到达终点时最少损失多少量。

思路：

d[u]表示到达该点时剩余量的最大值。

将松弛条件修改为大时更新即可。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 const int maxn = +;

 int n,m,tot,s,t,M;
 int head[maxn];
 double d[maxn];
 bool done[maxn];

 struct node
 {
     int v,next;
     double w;
 }e[*maxn];

 struct HeapNode
 {
     double d;
     int u;
     HeapNode(int u, double d):u(u),d(d){}
     bool operator< (const HeapNode& rhs) const
     {
         return d < rhs.d;
     }
 };

 void addEdge(int u, int v, double w)
 {
     e[tot].v = v;
     e[tot].w = w;
     e[tot].next = head[u];
     head[u] = tot++;
 }

 void dijkstra(int s)
 {
     priority_queue<HeapNode> Q;
     for(int i=;i<=n;i++)  d[i] = ;
     d[s] = M;
     memset(done,,sizeof(done));
     Q.push(HeapNode(s,M));
     while(!Q.empty())
     {
         HeapNode x = Q.top(); Q.pop();
         int u = x.u;
         if(done[u])  continue;
         done[u] = true;
         for(int i=head[u];i!=-;i=e[i].next)
         {
             int v = e[i].v;
             if(d[v] < d[u] - d[u]*e[i].w/100.0)
             {
                 d[v] = d[u] - d[u]*e[i].w/100.0;
                 Q.push(HeapNode(v,d[v]));
             }
         }
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     while(~scanf("%d",&n))
     {
         tot = ;
         memset(head,-,sizeof(head));
         for(int i=;i<=n;i++)
         {
             scanf("%d",&m);
             while(m--)
             {
                 int u;double w;
                 scanf("%d%lf",&u,&w);
                 addEdge(i,u,w);
             }
         }
         scanf("%d%d%d",&s,&t,&M);
         dijkstra(s);
         printf("%.2f\n",M-d[t]);
     }
     return ;
 }