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 = 50000+5;

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

struct node
{
    int v,next;
    double w;
}e[50*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=0;i<=n;i++)  d[i] = 0;
    d[s] = M;
    memset(done,0,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!=-1;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 = 0;
        memset(head,-1,sizeof(head));
        for(int i=1;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 0;
}