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 !!!