P2685 [TJOI2012]桥

P2685 [TJOI2012]桥

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 << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1e5 + ;
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 = ;
bool vis[N*];
struct edge {
    int to, w, nxt;
}edge[N*];
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<<], lazy[N<<];
void push_up(int rt) {
    tree[rt] = min(tree[rt<<], tree[rt<<|]);
}
void push_down(int rt) {
    tree[rt<<] = min(tree[rt<<], lazy[rt]);
    tree[rt<<|] = min(tree[rt<<|], lazy[rt]);
    lazy[rt<<] = min(lazy[rt<<], lazy[rt]);
    lazy[rt<<|] = min(lazy[rt<<|], 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 >> ;
    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 >> ;
    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 >> ;
    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 = ; i <= n; ++i) d[i] = INF;
    d[s] = ;
    q.push({, 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] ==  && id[v] ==  && d[u] + w == d[v]) dfs(v, s, bl, d);
    }
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = ; i <= m; ++i) {
        scanf("%d %d %d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }
    Dijkstra(, d1);
    Dijkstra(n, d2);
    tot = ;
    u = ;
    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+)/] = true;
                u = v;
                break;
            }
        }
    }
    link[++tot] = n;
    id[n] = tot;
    for (int i = ; i <= tot; ++i) dfs(link[i], link[i], l, d1);
    for (int i = tot; i >= ; --i) dfs(link[i], link[i], r, d2);
    build(, , tot-);
    for (int u = ; 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+)/]) {
                if(id[l[u]] < id[r[v]]) update(id[l[u]], id[r[v]]-, d1[u] + w + d2[v], , , tot-);
            }
        }
    }
    down(, , tot-);
    int ans = , cnt = ;
    for (int i = ; i < tot; ++i) {
        if(a[i] > ans) {
            ans = a[i];
            cnt = ;
        }
        else if(a[i] == ans) {
            cnt++;
        }
    }
    if(ans == d1[n]) cnt = m;
    printf("%d %d\n", ans, cnt);
    return ;
}