标签:flush solution main 无代码 return back vector auto clu
本题可以用最大流也可以用最大匹配(本质一样),用dinic最大流好建图,但码量大,匈牙利码量小,建图费点劲。
最大流:依旧是设一个源点一个汇点,对于每一个种类,连一条到汇点的边,capacity为需要的量,对于每一个试题,从源点连一条capacity为1的边到他,从他对每一个其所属的编号种类连一条capacity为1的边,求最大流即可,再找出最小割即可
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 1e5+5; const int maxn = 1e3+33; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, level[maxn], capacity[23]; vector<int> ans[23]; void init() { memset(head, -1, sizeof(head)); } void addedge(int u, int v, int cap) { edges[cnt] = edge{u, v, cap, 0, head[u]}; head[u] = cnt++; } void bfs(int s) { memset(level, -1, sizeof(level)); queue<int> q; level[s] = 0; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); for(int i = head[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; if(now.cap > now.flow && level[now.v] < 0) { level[now.v] = level[u] + 1; q.push(now.v); } } } } int dfs(int u, int t, int f) { if(u == t) return f; for(int& i = cur[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; if(now.cap > now.flow && level[u] < level[now.v]) { int d = dfs(now.v, t, min(f, now.cap - now.flow)); if(d > 0) { now.flow += d; edges[i^1].flow -= d; return d; } } } return 0; } int dinic(int s, int t) { int maxflow = 0; for(;;) { bfs(s); if(level[t] < 0) break; memcpy(cur, head, sizeof(head)); int f; while((f = dfs(s, t, INF)) > 0) maxflow += f; } return maxflow; } void run_case() { int k, n, p, u, sum = 0; init(); cin >> k >> n; int s = 0, t = n+k+1; for(int i = 1; i <= k; ++i) { cin >> capacity[i]; sum += capacity[i]; addedge(n+i, t, capacity[i]), addedge(t, n+i, 0); } for(int i = 1; i <= n; ++i) { cin >> p; addedge(s, i, 1), addedge(i, s, 0); while(p--) { cin >> u; addedge(i, u+n, 1), addedge(u+n, i, 0); } } if(dinic(s, t) != sum) {cout << "No Solution!"; return;} for(int i = 1; i <= n; ++i) { for(int j = head[i]; j != -1; j = edges[j].nex) { if(edges[j].flow) { ans[edges[j].v - n].push_back(i); break; } } } for(int i = 1; i <= k; ++i) { cout << i << ":"; for(auto j : ans[i]) cout << " " << j; cout << "\n"; } } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); //cout.flush(); return 0; }
匈牙利:对于每一个种类都连一条边到其所属的试卷,求最大匹配即可(无代码)
标签:flush solution main 无代码 return back vector auto clu
原文地址:https://www.cnblogs.com/GRedComeT/p/12274919.html