C F 715 B − C o m p l e t e T h e G r a p h \mathrm{CF715B - Complete\ The\ Graph} CF715B−Complete The Graph
D e s c r i p t i o n \mathrm{Description} Description
给定一张 n n n 个点, m m m 条边的无向图,点的编号为 0 ∼ n − 1 0\sim n-1 0∼n−1,对于每条边权为 0 0 0 的边赋一个不超过 1 0 18 10^{18} 1018 的正整数权值,使得 S S S 到 T T T 的最短路长度为 L L L。
S o l u t i o n \mathrm{Solution} Solution
W a y 1 \mathrm{Way\ 1} Way 1
考虑将每 1 1 1 条长度为 0 0 0 的边记录出来,初始将其全部设置为 1 1 1(因为要求边权值 ∈ [ 1 , 1 0 18 ] \in[1,10^{18}] ∈[1,1018]),如果将这些边依次不断地加 1 1 1,则 S S S 到 T T T 的最短路的长度会不断地增加或不变,总之最短路长度是单调不降的。那么,如果有解就必定会找到一种方案,反之则不会。
观察数据范围可知,最多每条边会加到 L L L,有 m m m 条边,那么时间应为 O ( m 2 L log n ) O(m^2L\log n) O(m2Llogn),因为还需加入 Dijkstra 的时间复杂度。
显然,会 TLE。不过,上文已分析最短路的长度是单调不降的,所以满足二分的性质,可以二分总共加 1 1 1 的个数,然后对于每跳边先加 ⌊ 个数 边数 ⌋ \lfloor\frac{个数}{边数}\rfloor ⌊边数个数⌋,之后对于 1 ∼ 个数 m o d 边数 1\sim 个数\bmod 边数 1∼个数mod边数 的边再加 1 1 1 即可。
时间复杂度: O ( m log L log n ) O(m\log L\log n) O(mlogLlogn)
C o d e Code Code
#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int SIZE = 2e4 + 10;int N, M, L, S, T;
int h[SIZE], e[SIZE], ne[SIZE], w[SIZE], idx;
std::vector<int> Id;
int Dist[SIZE], Vis[SIZE];void add(int a, int b, int c)
{e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}int Dijkstra()
{priority_queue<PII, vector<PII>, greater<PII>> Heap;memset(Dist, 0x3f, sizeof Dist);memset(Vis, 0, sizeof Vis);Dist[S] = 0, Heap.push({0, S});while (Heap.size()){auto Tmp = Heap.top();Heap.pop();int u = Tmp.second;if (Vis[u]) continue;Vis[u] = 1;for (int i = h[u]; ~i; i = ne[i]){int j = e[i];if (Dist[j] > Dist[u] + w[i]){Dist[j] = Dist[u] + w[i];Heap.push({Dist[j], j});}}}return Dist[T];
}int Check(int X)
{for (auto c : Id)w[c] = X / (int)(Id.size() / 2);for (int i = 0; i < (X % (int)(Id.size() / 2)) * 2; i += 2)w[Id[i]] += 1, w[Id[i] ^ 1] += 1;return Dijkstra();
}signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);memset(h, -1, sizeof h);cin >> N >> M >> L >> S >> T;int u, v, c, Cpy = M;while (M --){cin >> u >> v >> c;if (c) add(u, v, c), add(v, u, c);else{Id.push_back(idx), add(u, v, 1);Id.push_back(idx), add(v, u, 1);}}M = Cpy;if (!Id.size()){if (Dijkstra() == L){cout << "YES" << endl;for (int i = 0; i < idx; i += 2)cout << e[i ^ 1] << " " << e[i] << " " << w[i] << endl;}elsecout << "NO" << endl;return 0;}int l = Id.size() / 2, r = L * M;while (l < r){int mid = l + r >> 1;if (Check(mid) >= L) r = mid;else l = mid + 1;}if (Check(r) != L){cout << "NO" << endl;return 0;}cout << "YES" << endl;for (int i = 0; i < idx; i += 2)cout << e[i ^ 1] << " " << e[i] << " " << w[i] << endl;return 0;
}
W a y 2 \mathrm{Way\ 2} Way 2
从 S S S 开始先做 1 1 1 遍 Dijkstra,记当前 L L L 与 S S S 到 T T T 的最短路的差值为 D i f f Diff Diff(即 D i f f = L − D 1 , T Diff=L-D_{1,T} Diff=L−D1,T)
之后再做第 2 2 2 遍 Dijkstra 的时候,当点 u u u 更新至点 v v v 时且当前边为特殊边(初始变为 0 0 0),若 D 2 , u + w i < D 1 , v + D i f f D_{2,u}+w_i< D_{1,v}+Diff D2,u+wi<D1,v+Diff,则说明这时候最短路长度少了,尽量要让其补上这缺失的部分,即 w i = D 1 , u + D i f f − D 2 , u w_i = D_{1,u}+Diff-D_{2,u} wi=D1,u+Diff−D2,u。修改后,再进行正常 Dijkstra 的更新即可。
注:
D 1 , i D_{1,i} D1,i 表示第 1 1 1 次 Dijkstra 到达 i i i 号点的最短路长度, D 2 , i D_{2,i} D2,i 表示第 2 2 2 次 Dijkstra 到达第 i i i 号点的最短路长度。
C o d e Code Code
#include <bits/stdc++.h>
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int SIZE = 2e4 + 10;int N, M, L, S, T;
int h[SIZE], e[SIZE], ne[SIZE], w[SIZE], f[SIZE], idx;
int D[2][SIZE], Vis[SIZE];void add(int a, int b, int c)
{if (!c) f[idx] = 1;e[idx] = b, ne[idx] = h[a], w[idx] = max(1ll, c), h[a] = idx ++;
}void Dijkstra(int dist[], int Turn)
{for (int i = 0; i < N; i ++)dist[i] = 1e18, Vis[i] = 0;priority_queue<PII, vector<PII>, greater<PII>> Heap;Heap.push({0, S}), dist[S] = 0;while (Heap.size()){auto Tmp = Heap.top();Heap.pop();int u = Tmp.second;if (Vis[u]) continue;Vis[u] = 1;for (int i = h[u]; ~i; i = ne[i]){int j = e[i];if (Turn == 2 && f[i] && dist[u] + w[i] < D[0][j] + L - D[0][T])w[i] = w[i ^ 1] = D[0][j] + L - D[0][T] - dist[u];if (dist[j] > dist[u] + w[i]){dist[j] = dist[u] + w[i];Heap.push({dist[j], j});}}}
}signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);memset(h, -1, sizeof h);cin >> N >> M >> L >> S >> T;int u, v, c;while (M --){cin >> u >> v >> c;add(u, v, c), add(v, u, c);}Dijkstra(D[0], 1);if (L - D[0][T] < 0){cout << "NO" << endl;return 0;}Dijkstra(D[1], 2);if (D[1][T] != L){cout << "NO" << endl;return 0;}cout << "YES" << endl;for (int i = 0; i < idx; i += 2)cout << e[i ^ 1] << " " << e[i] << " " << w[i] << endl;return 0;
}