題意略。

思路：為了保證每個點都有至少k條邊覆蓋，我們可以讓二分圖的左半邊與源點s相連，連容量為indegree[i] - k的邊（如果正著想不好想，我們可以想它的反面，

限制它反面的上限，從而保證我正面k條邊的覆蓋），讓二分圖的右半邊與匯點t相連，容量同樣是indegree[i] - k。然后跑最大流，E - 最大流中用到的邊 = 答案。

每次跑完最大流都要重新建圖。我的最大流用的是dinic。

詳見代碼：

#include<bits/stdc++.h> #define maxn 4005 #define INF 0x3f3f3f3f using namespace std;struct edge{int to,cap,rev,id;edge(int a = 0,int b = 0,int c = 0,int d = 0){to = a,cap = b,rev = c,id = d;} };bool visit[maxn]; int indg[maxn],level[maxn],iter[maxn],n1,n2,m,s,t,store[maxn]; vector<edge> graph[maxn]; vector<edge> standard[maxn];void add_e(int from,int to,int cap,int id){graph[from].push_back(edge(to,cap,graph[to].size(),id));graph[to].push_back(edge(from,0,graph[from].size() - 1,id)); } void bfs(int s){memset(level,-1,sizeof(level));queue<int> que;level[s] = 0;que.push(s);while(que.size()){int v = que.front();que.pop();for(int i = 0;i < graph[v].size();++i){edge& e = graph[v][i];if(e.cap > 0 && level[e.to] < 0){level[e.to] = level[v] + 1;que.push(e.to);}}} } int dfs(int v,int t,int f){if(v == t) return f;for(int& i = iter[v];i < graph[v].size();++i){edge& e = graph[v][i];if(e.cap > 0 && level[v] < level[e.to]){int d = dfs(e.to,t,min(f,e.cap));if(d > 0){e.cap -= d;graph[e.to][e.rev].cap += d;return d;}}}return 0; } int max_flow(int s,int t){int flow = 0;while(true){bfs(s);if(level[t] < 0) return flow;memset(iter,0,sizeof(iter));int f;while((f = dfs(s,t,INF)) > 0){flow += f;}} }int main(){scanf("%d%d%d",&n1,&n2,&m);s = 0,t = n1 + n2 + 1;for(int i = 0;i < m;++i){int u,v;scanf("%d%d",&u,&v);v += n1;++indg[u],++indg[v];standard[u].push_back(edge(v,1,standard[v].size(),i + 1));standard[v].push_back(edge(u,0,standard[u].size() - 1,i + 1));}int mdg = maxn;for(int i = 1;i <= n1 + n2;++i)mdg = min(mdg,indg[i]);printf("0\n");for(int k = 1;k <= mdg;++k){for(int v = 0;v < maxn;++v) graph[v].clear();for(int i = 1;i <= n1 + n2;++i){for(int j = 0;j < standard[i].size();++j)graph[i].push_back(standard[i][j]);}for(int i = 1;i <= n1;++i) add_e(s,i,indg[i] - k,-1);for(int i = n1 + 1;i <= n1 + n2;++i) add_e(i,t,indg[i] - k,-1);max_flow(s,t);memset(visit,false,sizeof(visit));int cnt = 0;for(int i = n1 + 1;i <= n1 + n2;++i){for(int j = 0;j < graph[i].size();++j){edge& e = graph[i][j];if(e.cap && e.to != t){visit[e.id] = true;++cnt;}}}printf("%d",m - cnt);for(int i = 1;i <= m;++i){if(!visit[i]) printf(" %d",i);}printf("\n");}return 0; }/* 1 2 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 */

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轉載于:https://www.cnblogs.com/tiberius/p/9032098.html

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