acwing1126. 最小花费

题意

思路

从终点 e 向起点 s 倒着跑 spfa，把 spfa 在根据题意稍微改一改。

代码

#include <bits/stdc++.h>
using namespace std;
#define db  double
#define ll  long long
#define Pir pair<int, int>
#define fi  first
#define se  second
#define pb  push_back
#define m_p make_pair
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
/*==========ACMer===========*/
const int N = 2005, M = 2e5 + 10;
int n, m, s, t;
struct Edge
{
    
    
    int v, w, ne;
} e[M];
int h[N], tot;
void add(int u, int v, int w)
{
    
    
    e[++ tot] = {
    
     v, w, h[u] }; h[u] = tot;
}

bool vis[N];
db dis[N];


void spfa()
{
    
    
    //0x7f double 正无穷 !!!
    //0x8f double 负无穷 !!!
    memset(dis, 0x7f, sizeof dis);
    dis[s] = 100, vis[s] = 1;
    queue<int> q;
    q.push(s);
    while (q.size())
    {
    
    
        int u = q.front(); q.pop();
        vis[u] = 0;
        for (int i = h[u]; i; i = e[i].ne) {
    
    
            int v = e[i].v, w = e[i].w;
            if (dis[v] > dis[u] / (1.0 - w / 100.0))
            {
    
    
                dis[v] = dis[u] / (1.0 - w / 100.0);
                if (! vis[v])
                {
    
    
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}


int main()
{
    
    
    scanf("%d %d", &n, &m);
    int u, v, w;
    for (int i = 0; i < m; i ++) {
    
    
        scanf("%d %d %d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }

    scanf("%d %d", &t, &s);
    spfa();
    printf("%.8f\n", dis[t]);


    return 0;
}