思路
这道题乍一看像是最短路，但又不知道从何下手。
考虑枚举每个节点，假设两个人到第 (i) 个节点相遇，剩下的路程一块走，那么我们很容易得出当前的能量为 (dis_{1~to~i} imes B + dis_{2~to~i} imes E + dis_{i~to~n} * P.)
那么我们可以预处理跑三遍最短路。
然后枚举每个节点寻找最优解。
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 4e4 + 10;
queue <int > s;
struct node {
int next ,to;
}edge[MAXN << 1];
int head[MAXN] ,cnt = 0;
void add (int from ,int to) {
edge[++ cnt].next = head[from];
head[from] = cnt;
edge[cnt].to = to;
}
int B ,E ,P ,n ,m;
int d[3][MAXN];
void bfs (int st) {
int vis[MAXN] ,a[MAXN];
memset (a ,-1 ,sizeof (a));
s.push(st);
a[st] = 0;
while (! s.empty()) {
int u = s.front();
s.pop();
for (int q = head[u];~ q;q = edge[q].next) {
int v = edge[q].to;
if (a[v] == -1) {
a[v] = a[u] + 1;
s.push(v);
}
}
}
if (st != n)
for (int q = 1;q <= n;++ q)
d[st][q] = a[q];
else
for (int q = 1;q <= n;++ q)
d[0][q] = a[q];
return ;
}
int ans = 0x3f3f3f3f;
int main () {
memset (head ,-1 ,sizeof (head));
scanf ("%d%d%d%d%d",&B ,&E ,&P ,&n ,&m);
int from ,to;
for (int q = 1;q <= m;++ q) {
scanf ("%d%d",&from ,&to);
add (from ,to) ,add (to ,from);
}
bfs (1); bfs (2); bfs (n);
for (int q = 1;q <= n;++ q) {
ans = min (ans ,d[1][q] * B + d[2][q] * E + d[0][q] * P);
}
printf ("%d
",ans);
return 0;
}