HDU 4318 Power transmission (dijkstra && 邻接表从队友那里抄来的，当作模板吧)

本文链接：https://blog.youkuaiyun.com/u010557442/article/details/12260123

本文介绍了一种基于图的最短路径算法实现，通过使用邻接表存储边信息，并结合松弛操作更新节点间的最短距离。该算法适用于解决带有权重的有向图中的最短路径问题。

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#pragma comment(linker, "/STACK:102400000,102400000")
#include "iostream"
#include "cstring"
#include "algorithm"
#include "cmath"
#include "cstdio"
#include "sstream"
#include "queue"
#include "vector"
#include "string"
#include "stack"
#include "cstdlib"
#include "deque"
#include "fstream"
#include "map"
using namespace std;
typedef long long LL;
const int INF = 0x1fffffff;
const int MAXN = 1000000+100;
#define eps 1e-14
const int mod = 100000007;

int head[MAXN],vis[MAXN];
int cnt,n;
double dis,low[MAXN];

struct edge
{
    int now;
    int next;
    double value;
}es[MAXN];

void addedge(int a,int b,int c)
{
    es[cnt].now=b;
    es[cnt].next=head[a];
    es[cnt].value=0.01*c;
    head[a]=cnt++;
}

double dij(int s,int t,int w)
{
    for (int i=0;i<=n;i++)
    {
        low[i]=-1;
        vis[i]=0;
    }
    low[s]=w*1.0;
    for (int i=1;i<n;i++)
    {
        int pos=-1;
        for (int i=1;i<=n;i++)
            if (!vis[i]&&(pos==-1||low[i]>low[pos]))
            pos=i;
        if (pos==t) break;
        vis[pos]=1;
        for (int i=head[pos];i!=-1;i=es[i].next)
        if(!vis[es[i].now]&&(low[es[i].now]<low[pos]*(1-es[i].value)))
        low[es[i].now]=low[pos]*(1-es[i].value);
    }
    return low[t];
}
int main()
{
    freopen("in","r",stdin);
    while (scanf("%d",&n)!=EOF)
    {
        dis=-1;
        cnt=1;
        memset(head,-1,sizeof head);
        for (int i=1;i<=n;i++)
        {
            int k;
            scanf("%d",&k);
            while (k--)
            {
                int b,c;
                scanf("%d%d",&b,&c);
                addedge(i,b,c);
            }


        }
        int s,t,w;
        scanf("%d%d%d",&s,&t,&w);
        dis=dij(s,t,w);
        if (dis<0) printf("IMPOSSIBLE!\n");
        else printf("%0.2lf\n" , (1.0*w - dis) );
    }
  return 0;
}