一、内容
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field nat the end of the day.The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has kspecial fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.After the road is added, Bessie will return home on the shortest path from field 1to field n . Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2≤n≤2⋅105, n−1≤m≤2⋅105, 2≤k≤n) — the number of fields on the farm, the number of roads, and the number of special fields.The second line contains kintegers a1,a2,…,ak (1≤ai≤n) — the special fields. All aiare distinct.The i-th of the following m lines contains integers xi and yi (1≤xi,yi≤n, xi≠yi), representing a bidirectional road between fields xi and yi It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1to nafter Farmer John installs one road optimally.
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
二、思路
三、代码
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int N = 2e5 + 5, M = N * 2;
struct E {int v, next;} e[M];
int n, m, k, u, v, p[N], len, h[N], d1[N], d2[N];
void add(int u, int v) {e[++len].v = v; e[len].next = h[u]; h[u] = len;}
void bfs(int d[], int u) {
queue<int> q; q.push(u);
memset(d, -1, sizeof(d1)); d[u] = 0;
while (!q.empty()) {
u = q.front(); q.pop();
for (int j = h[u]; j; j = e[j].next) {
int v = e[j].v;
if (d[v] == -1) d[v] = d[u] + 1, q.push(v);
}
}
}
//按照 d1[u] - d2[u]从小到大排序 这样排序后的点(d1[u] - d2[u]) < (d1[v] - d2[v])
bool cmp(int u, int v) {return (d1[u] - d2[u]) < (d1[v] - d2[v]);}
int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= k; i++) scanf("%d", &p[i]);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &u, &v); add(u, v); add(v, u);
}
bfs(d1, 1);//1到其他点的最短路
bfs(d2, n);//n到其他点的最短路
sort(p + 1, p + 1 + k, cmp);
int ans = 0, mx = -1e9; //由于第一个之前没有点 所以mx为负无穷
for (int i = 1; i <= k; i++) {
v = p[i];
//从前往后遍历 这样保证 (d1[u] - d2[u]) < (d1[v] - d2[v])
//那么我们只需要加上1~v-1之前最大的d1 mx即可。
ans = max(ans, mx + d2[v]);
mx = max(mx, d1[v]);
}
printf("%d", min(ans + 1, d1[n]));
return 0;
}
