"蔚来杯"2022牛客暑期多校训练营7 L Maximum Range（强连通缩点+网络流输出方案）

题意

找一个环上面的边权极差最大并输出点

思路

我们先强联通缩点统计出每个强连通分量里的边
难点是在输出方案找环
我们将源点连向最大边权的两个端点最小边权的两个端点
我们在跑最大流的时候记录每条边的流量那么我们最终的答案就是最大边和最小边以及有流量的边

代码巨多细节 code挺难的但偏板子也挺好写的

#include <bits/stdc++.h>
using namespace  std;
//#define  int long long
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
#define fi first
#define se second
#define pb  push_back
#define inf 1e18
#define de_bug(x) cerr << #x << "=" << x << endl
#define all(a) (a).begin(),(a).end()
#define IOS   std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define  fer(i,a,b)  for(int i=a;i<=b;i++)
#define  der(i,a,b)  for(int i=a;i>=b;i--)
const int mod = 1e9 + 7;
const int N = 1e6 + 10;const int V = 1e6 + 10;
const int E = 1e6 + 10;
template<typename T>
struct FlowGraph {int s, t, vtot;int h[V], idx;int d[V], cur[V];int e[E * 2];int ne[E * 2];int f[E * 2];int  ff[E * 2];/*struct edge {int v, nxt;T f;} e[E * 2];void addedge(int u,int v, T f){e[etot]= {v, head[u], f}; head[u] = etot++;e[etot]= {u, head[v], 0}; head[v] = etot++;}*/void add(int a, int b, int c) {e[idx] = b, f[idx] = c, ne[idx] = h[a], h[a] = idx++;e[idx] = a, f[idx] = 0, ne[idx] = h[b], h[b] = idx++;}void add1(int a, int b, int c) {e[idx] = b, f[idx] = c, ne[idx] = h[a], h[a] = idx++;e[idx] = a, f[idx] = c, ne[idx] = h[b], h[b] = idx++;}bool bfs() {for (int i = 1; i <= vtot; i++) {d[i] = 0;cur[i] = h[i];}queue<int> q;q.push(s);d[s] = 1;while (!q.empty()) {int u = q.front();q.pop();for (int i = h[u]; ~i; i = ne[i]) {int v = e[i];if (f[i] && !d[v]) {d[v] = d[u] + 1;if (v == t) return true;q.push(v);}}}return false;}/*   T dfs(int u, T m) {if (u == t) return m;T flow = 0;for (int i = cur[u]; ~i; cur[u] = i = ne[i]){int v=e[i];if (f[i] && d[v] == d[u] + 1) {T t = dfs(v, min(m-flow, f[i]));if(!t)d[v]=-1;f[i] -= t;f[i ^ 1] += t;flow+=t;}}return flow;}*/T dfs(int u, T m) {if (u == t) return m;T flow = 0;for (int i = cur[u]; ~i; cur[u] = i = ne[i])if (f[i] && d[e[i]] == d[u] + 1) {T t = dfs(e[i], min(m, f[i]));f[i] -= t;f[i ^ 1] += t;ff[i] += t;ff[i ^ 1] -= t;m -= t;flow += t;if (!m) break;}if (!flow) d[u] = -1;return flow;}T dinic() {T flow = 0;while (bfs()) flow += dfs(s, numeric_limits<T>::max());return flow;}void init(int s_, int t_, int vtot_) {s = s_;t = t_;vtot = vtot_;idx = 0;for (int i = 1; i <= vtot; i++) h[i] = -1;}void dfs1(int u, vector<int>&ans) {ans.push_back(u);for(int i = h[u]; ~i; i = ne[i]) {int v = e[i];if(ff[i] > 0) {--ff[i];dfs1(v, ans);break;}}}
};
FlowGraph<int> gg;
int n, m, s, t;
const int M = 1e6 + 100;pii ve[M];
vector<pii> g[N];
int dfn[N], low[N], idx, cnt;
vector<int>cc[N];
int bridge[M];
int e[M];
int vis[N];
void tarjan(int u, int id) {dfn[u] = low[u] = ++idx;for ( pii t : g[u]) {int  v = t.fi;int id2 = t.se;if (!dfn[v]) {tarjan(v, id2);low[u] = min(low[u], low[v]);if (dfn[v] == low[v]) bridge[id2] = 1;} else if (id != id2) {low[u] = min(low[u], dfn[v]);}}
}
void dfs(int u) {vis[u] = 1;for(pii t : g[u]) {int v = t.fi;int id = t.se;if(!bridge[id]) {cc[cnt].push_back(id);if(!vis[v])dfs(v);}}
}
bool cmp(int x, int y) {return e[x] < e[y];
}
void solve() {cin >> n >> m;fer(i, 1, m) {int a, b;cin >> a >> b >> e[i];g[a].pb({b, i});g[b].pb({a, i});ve[i] = {a, b};}tarjan(1, 0);fer(i, 1, n) {if(!vis[i])++cnt, dfs(i);}int maa = -1, res = 0;fer(i, 1, cnt) {sort(all(cc[i]));cc[i].erase(unique(all(cc[i])), cc[i].end());sort(all(cc[i]), cmp);if(cc[i].size() >= 2) {int res1 = e[cc[i].back()] - e[*cc[i].begin()];if(res1 > maa) {maa = res1;res = i;}}}cout << maa << endl;vector<int>ans1;vector<int>ans2;s = n + 1, t = n + 2;int ma = cc[res].back();int mi = *cc[res].begin();gg.init(s, t, t);gg.add(s, ve[ma].fi, 1);gg.add(s, ve[ma].se, 1);gg.add(ve[mi].fi, t, 1);gg.add(ve[mi].se, t, 1);for(int x : cc[res]) {if(x != ma && x != mi) {gg.add1(ve[x].fi, ve[x].se, 1);}}gg.dinic();gg.dfs1(s, ans1);gg.dfs1(s, ans2);cout << ans1.size() + ans2.size() - 4 << endl;for(int x : ans1) {if(x != s && x != t) cout << x << " ";}reverse(ans2.begin(), ans2.end());for(int x : ans2) {if(x != s && x != t) cout << x << " ";}cout << endl;
}
int main() {IOS;int _ = 1;//cin>>_;while( _-- )solve();
}

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“蔚来杯“2022牛客暑期多校训练营7 L Maximum Range（强连通缩点+网络流输出方案）

"蔚来杯"2022牛客暑期多校训练营7 L Maximum Range（强连通缩点+网络流输出方案）

题意

思路

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