以 1，2，n 为起点跑3次 bfs 或者 spfa
那么 ans = min(ans, dis[1][i] * B + dis[2][i] * E + dis[3][i] * P) (1 <= i <= n)
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#define N 40001
#define LL long long
#define min(x, y) ((x) < (y) ? (x) : (y))
int B, E, P, n, m, cnt;
int dis[4][N], head[N], to[N << 1], next[N << 1];
bool vis[N];
std::queue <int> q;
LL ans = ~(1 << 31);
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
return x * f;
}
inline void add(int x, int y)
{
to[cnt] = y;
next[cnt] = head[x];
head[x] = cnt++;
}
inline void spfa(int s, int k)
{
int i, v, u;
memset(vis, 0, sizeof(vis));
memset(dis[k], 127, sizeof(dis[k]));
q.push(s);
dis[k][s] = 0;
while(!q.empty())
{
u = q.front();
q.pop();
vis[u] = 0;
for(i = head[u]; i ^ -1; i = next[i])
{
v = to[i];
if(dis[k][v] > dis[k][u] + 1)
{
dis[k][v] = dis[k][u] + 1;
if(!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
}
int main()
{
int i, x, y;
B = read();
E = read();
P = read();
n = read();
m = read();
memset(head, -1, sizeof(head));
for(i = 1; i <= m; i++)
{
x = read();
y = read();
add(x, y);
add(y, x);
}
spfa(1, 1);
spfa(2, 2);
spfa(n, 3);
for(i = 1; i <= n; i++)
ans = min(ans, (LL)(dis[1][i] * B + dis[2][i] * E + dis[3][i] * P));
printf("%lld
", ans);
return 0;
}