思路：
先求出最短路： d1[u] : u 到 1 的最短路， d2[u] : u 到 n 的最短路
再求出一条从 1 到 n 的最短路链，然后从链上的每一个点出发dfs, 求出：
l[u] : u 到 1 的最短路径过中和链的交点（离 1 最近的）
r[u] : u 到 n 的最短路径过中和链的交点（离 n 最近的）
然后对于一条非链上的边（ u -> v ），边权为 w ，对于链上的 l[u] 到 r[v] 之间的边任意删一条边，
最短路都有可能变成 d1[u] + w + d2[v] 然后在链上维护这个的最小值，就能知道删掉链上的每条边对最短路的影响
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << " "; #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //head const int N = 1e5 + 5; const int INF = 0x3f3f3f3f; int n, m, u, v, w, tot, d1[N], d2[N], link[N], id[N], l[N], r[N], a[N]; int head[N], cnt = 0; bool vis[N*2]; struct edge { int to, w, nxt; }edge[N*4]; void add(int u, int v, int w) { edge[++cnt] = {v, w, head[u]}; head[u] = cnt; } priority_queue<pii, vector<pii>, greater<pii> > q; int tree[N<<2], lazy[N<<2]; void push_up(int rt) { tree[rt] = min(tree[rt<<1], tree[rt<<1|1]); } void push_down(int rt) { tree[rt<<1] = min(tree[rt<<1], lazy[rt]); tree[rt<<1|1] = min(tree[rt<<1|1], lazy[rt]); lazy[rt<<1] = min(lazy[rt<<1], lazy[rt]); lazy[rt<<1|1] = min(lazy[rt<<1|1], lazy[rt]); lazy[rt] = INF; } void build(int rt, int l, int r) { lazy[rt] = INF; if(l == r) { tree[rt] = INF; return ; } int m = l+r >> 1; build(ls); build(rs); push_up(rt); } void down(int rt, int l, int r) { if(l == r) { a[l] = tree[rt]; return ; } if(lazy[rt] != INF) push_down(rt); int m = l+r >> 1; down(ls); down(rs); push_up(rt); } void update(int L, int R, int x, int rt, int l, int r) { if(L <= l && r <= R) { tree[rt] = min(tree[rt], x); lazy[rt] = min(lazy[rt], x); return ; } if(lazy[rt] != INF) push_down(rt); int m = l+r >> 1; if(L <= m) update(L, R, x, ls); if(R > m) update(L, R, x, rs); push_up(rt); } void Dijkstra(int s, int *d) { for (int i = 1; i <= n; ++i) d[i] = INF; d[s] = 0; q.push({0, s}); while(!q.empty()) { pii p = q.top(); q.pop(); int u = p.se; if(d[u] < p.fi) continue; for (int i = head[u]; i; i = edge[i].nxt) { int v = edge[i].to; int w = edge[i].w; if(d[v] > p.fi + w) { d[v] = p.fi + w; q.push({d[v], v}); } } } } void dfs(int u, int s, int *bl, int *d) { bl[u] = s; for (int i = head[u]; i; i = edge[i].nxt) { int v = edge[i].to; int w = edge[i].w; //if(vis[(i+1)/2]) continue; if(bl[v] == 0 && id[v] == 0 && d[u] + w == d[v]) dfs(v, s, bl, d); } } int main() { scanf("%d %d", &n, &m); for (int i = 1; i <= m; ++i) { scanf("%d %d %d", &u, &v, &w); add(u, v, w); add(v, u, w); } Dijkstra(1, d1); Dijkstra(n, d2); tot = 0; u = 1; while(u != n) { link[++tot] = u; id[u] = tot; for (int i = head[u]; i; i = edge[i].nxt) { int v = edge[i].to; int w = edge[i].w; if(d2[v] + w == d2[u]){ vis[(i+1)/2] = true; u = v; break; } } } link[++tot] = n; id[n] = tot; for (int i = 1; i <= tot; ++i) dfs(link[i], link[i], l, d1); for (int i = tot; i >= 1; --i) dfs(link[i], link[i], r, d2); build(1, 1, tot-1); for (int u = 1; u <= n; ++u) { for (int i = head[u]; i; i = edge[i].nxt) { int v = edge[i].to; int w = edge[i].w; if(!vis[(i+1)/2]) { if(id[l[u]] < id[r[v]]) update(id[l[u]], id[r[v]]-1, d1[u] + w + d2[v], 1, 1, tot-1); } } } down(1, 1, tot-1); int ans = 0, cnt = 0; for (int i = 1; i < tot; ++i) { if(a[i] > ans) { ans = a[i]; cnt = 1; } else if(a[i] == ans) { cnt++; } } if(ans == d1[n]) cnt = m; printf("%d %d ", ans, cnt); return 0; }