题目链接:最小花费
A 转账给 B 手续费 为 x1 , B 转账给 C 手续费为 x2 ,则由 A 转账给 C 手续费为 x1 * x2;
if(dis[to] < dis[now] * e[i].w)
dis[to] = dis[now] * e[i].w;
代码如下
#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>
using namespace std;
const int N = 101000;
const int maxn = 2018;
struct Node{
int ne;
int to;
double w;
}e[N<<1];
int head[maxn];
double dis[maxn];
int n,m,x,y,st,en,val,cnt;
bool vis[maxn];
void init()
{
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
cnt = 0;
}
void add(int u,int v,double val)
{
e[cnt].to = v;
e[cnt].ne = head[u];
e[cnt].w = 1 - val; // 剩余的费用
head[u] = cnt ++;
}
void SPFA()
{
queue<int>q;
q.push(st);
memset(dis,0,sizeof(dis));
vis[st] = 1;
dis[st] = 1;
while(!q.empty())
{
int now = q.front();
q.pop();
for(int i=head[now];~i;i=e[i].ne)
{
int to = e[i].to;
if(dis[to] < dis[now] * e[i].w )
{
dis[to] = dis[now] * e[i].w;
if(!vis[to])
{
q.push(to);
vis[to] = 1;
}
}
}
vis[now] = 0;
}
}
int main()
{
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&x,&y,&val);
add(x,y,double(val)/100);
add(y,x,double(val)/100);
}
scanf("%d%d",&st,&en);
SPFA();
double ans = 100/dis[en];
printf("%.8lf\n",ans);
return 0;
}
注意初始化 和 double !!!