[網路流24題] P2762 太空飛行計劃
阿新 • • 發佈：2021-11-18
最大權閉合子圖，需要方案
https://www.luogu.com.cn/problem/P2762
沒啥好說的，裸的最大權閉合子圖。
需要注意的是最後是否連通可以用 dep[v]判斷
點選檢視程式碼
#include <bits/stdc++.h> #define endl '\n' #define IOS \ ios::sync_with_stdio(0); \ cin.tie(0); \ cout.tie(0) #define P pair<int, int> typedef long long ll; using namespace std; const int N = 100 + 5; const int M = N * N + 5; const int INF = 0x3f3f3f3f; struct edge { int v, w, to; } e[M * 2]; int pre[N], cnt_edge, dep[N]; int S, T, z, head[N], sum; int n, m, q[N], cur[N]; void add(int u, int v, int w) { e[cnt_edge] = {v, w, head[u]}; head[u] = cnt_edge++; e[cnt_edge] = {u, 0, head[v]}; head[v] = cnt_edge++; } bool bfs() { for (int i = 0; i <= T; i++) dep[i] = 0; dep[S] = 1; int l = 0, r = 1; q[r] = S; while (l < r) { int u = q[++l]; for (int i = head[u]; i != -1; i = e[i].to) { int v = e[i].v; if (!dep[v] && e[i].w) dep[v] = dep[u] + 1, q[++r] = v; } } return dep[T]; } int dfs(int u, int mi) { int res = 0; if (mi == 0 || u == T) return mi; for (int &i = cur[u]; i != -1; i = e[i].to) { int v = e[i].v; if (dep[u] + 1 == dep[v] && e[i].w) { int minn = dfs(v, min(mi - res, e[i].w)); e[i].w -= minn; e[i ^ 1].w += minn; res += minn; if (res == mi) return res; } } if (res == 0) dep[u] = 0; return res; } int dinic() { ll res = 0; while (bfs()) { memcpy(cur, head, sizeof(head)); // cout<<res<<endl; res += dfs(S, INF); } return res; } int id(int x, int on) { return on * n + x; } int vals[105]; int main() { int m; cin >> n >> m; memset(head, -1, sizeof head); S = n + m + 1, T = S + 1; int val; ll sum = 0; for (int i = 1; i <= n; i++) { cin >> val; val = val > 0 ? val : -val; vals[i] = val; sum += val; add(S, id(i, 0), val); char tools[10000]; memset(tools, 0, sizeof tools); cin.getline(tools, 10000); int ulen = 0, tool; while (sscanf(tools + ulen, "%d", &tool) == 1) //之前已經用scanf讀完了贊助商同意支付該實驗的費用 { // tool是該實驗所需儀器的其中一個 //這一行，你可以將讀進來的編號進行儲存、處理，如連邊。 // cout << "add: " << i << " " << tool << endl; add(id(i, 0), id(tool, 1), INF); if (tool == 0) ulen++; else { while (tool) { tool /= 10; ulen++; } } ulen++; } } for (int i = 1; i <= m; i++) { cin >> val; val = val > 0 ? val : -val; vals[i + n] = val; add(id(i, 1), T, val); } int ans = sum - dinic(); // puts("-----------"); for (int i = 1; i <= n; i++) { if (dep[i]) cout << i << " "; } cout << endl; for (int i = 1; i <= m; i++) { if (dep[i + n]) cout << i << " "; } cout << endl; cout << ans << endl; return 0; }
放一下我最大權閉合子圖的筆記
• 有一個有向圖，每一個點都有一個權值（可以為正或負或0），選擇一個權值和最大的子圖，使得每個點的後繼都在子圖裡面，這個子圖就叫最大權閉合子圖。
• 解決方法：
• 從源點s向每個正權點連一條容量為權值的邊，每個負權點向匯點t連一條容量為權值的絕對值的邊，原邊容量全部為無限大。
• 求它的最小割，割掉後，與源點s連通的點構成最大權閉合子圖，權值為正權值之和-最小割。
PS 用 \(Dinic\) 來寫，是否與源點s連通可以用分層時的 \(dep\) 判斷，其他寫法可以用 \(dist\) 等
for (int i = 1; i <= n; i++) { if (dep[i]) cout << i << " ";//連通則輸出 }
如：https://nanti.jisuanke.com/t/41096
點選檢視程式碼
#include<iostream> #include <cstring> #include <cmath> #include <map> #include <vector> #include <algorithm> #define int ll using namespace std; typedef long long ll; const int M = 4000 + 5; const int N = 8e2 + 5; const int INF = 0x3f3f3f3f; int n1,n2,tot; struct node { int v,w,to; } edge[M*2]; int pre[N],cnt,dep[N]; int S,T,z,head[N],sum,id; int n,m,q[N],cur[N]; void add(int u,int v,int w) { edge[cnt]= {v,w,head[u]}; head[u]=cnt++; edge[cnt]= {u,0,head[v]}; head[v]=cnt++; } bool bfs() { for(int i=0; i<=T; i++) dep[i]=0; dep[S]=1; int l=0,r=1; q[r]=S; while(l<r) { int u=q[++l]; for(int i=head[u]; i!=-1; i=edge[i].to) { int v=edge[i].v; if(!dep[v]&&edge[i].w) dep[v]=dep[u]+1,q[++r]=v; } } return dep[T]; } int dfs(int u,int mi) { int res=0; if(mi==0||u==T) return mi; for(int &i=cur[u]; i!=-1; i=edge[i].to) { int v=edge[i].v; if(dep[u]+1==dep[v]&&edge[i].w) { int minn=dfs(v,min(mi-res,edge[i].w)); edge[i].w-=minn; edge[i^1].w+=minn; res+=minn; if(res==mi) return res; } } if(res==0) dep[u]=0; return res; } int dinic() { ll res=0; while(bfs()) { memcpy(cur,head,sizeof(head)); // cout<<res<<endl; res+=dfs(S,INF); } return res; } signed main(){ int t;cin >> t; while(t--){ cin >> n1 >> n2; S = 0;T = n1 + n2 + 1;cnt= 0; ll totp = 0; for(int i = 0; i <= n1 + n2 + 1; i++) head[i] = -1; for(int i = 1; i <= n1; i++){ int w;cin >> w; totp += w; add(S,i,w); } for(int i = 1; i <= n2; i++){ int w;cin >> w; add(i+n1,T,w); } int m1,m2,v; for(int u = 1; u <= n1; u++){ cin >> m1 >> m2; for(int i = 0; i < m1; i++){ cin >> v; add(u,v+n1,INF); } for(int i = 0; i < m2; i++){ cin >> v; add(u,v,INF); } } int ans = dinic(); cout<<totp - ans<<endl; } }