标签：head cli 覆盖 ace owb als include 多少 节点
题目中没有说球的上限是多少，只告诉了柱子，那么我们就应该以柱子为界去增加球，考虑将每两个能组成完全平方数的点连边，就形成了一个DAG(有向无环图)，由于是DAG，可以转换为最小覆盖问题，即最多有n条路径(柱子数)，求其能覆盖的最大点数，最小覆盖路径 = 节点数 - 最大匹配数，可以将其拆成二分图跑匈牙利/最大流，由Hall定理，|S| <= |T|，此处的|S|就等于节点数-最大匹配数，而|T|等于最小覆盖路径，就是柱子数n(n个路径必有n个节点)，在满足条件的情况下增加球的数量即可。
在求最小覆盖路径时，可以用匈牙利/dinic,首先都要先拆分成二分图，每个点拆成一个入点一个出点，用dinic时，就要再设一个源点一个汇点，进行多次增广路，每一次的flow就是其增加节点后增加的匹配数，而用匈牙利时可以不实际拆图，对每个节点都跑一次匈牙利即可。
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 1e5+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, nex;//, cap, flow, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, level[maxm], pre[maxm]; bool vis[maxm]; void init() { memset(head, -1, sizeof(head));memset(pre, -1, sizeof(pre)); } void addedge(int u, int v, int cap) { edges[cnt] = edge{u, v, head[u]};//cap, 0, head[u]}; head[u] = cnt++; } bool dfs(int u) { for(int i = head[u]; i != -1; i = edges[i].nex) { int v = edges[i].v; if(!vis[v]) { vis[v] = true; if(pre[v] == -1|| dfs(pre[v])) { pre[v] = u; return true; } } } return false; } void run_case() { init(); int n; cin >> n; int flow = 0, num = 0; while(num - flow <= n) { num++; for(int i = sqrt(num)+1; i*i<(num<<1); ++i) { //和dinic一样 由新加的点向原点连，保证能更新答案，因为从新点跑匈牙利 addedge(num, i*i-num,0); } flow += dfs(num); memset(vis, 0, sizeof(vis)); } cout << --num; for(int i = 1; i <= num; ++i) { if(!vis[i]) { cout << "\n"; for(int u = i; u!=-1; u = pre[u]) { vis[u] = true; cout << u << " "; } } } } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); //cout.flush(); return 0; }
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 1e5+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, level[maxm], pre[maxm]; bool vis[maxm]; 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; pre[u>>1] = now.v>>1; 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() { init(); int n; cin >> n; int s = 0, t = 1e5; int flow = 0, num = 0; while(num - flow <= n) { num++; addedge(s, num<<1, 1), addedge(num<<1, s, 0); addedge((num<<1)|1, t, 1), addedge(t, (num<<1)|1, 0); for(int i = sqrt(num)+1; i*i<(num<<1); ++i) { //是将原有的球的入点向新加的球的出点连边，保证能增加流 addedge((i*i-num)<<1, (num<<1)|1, 1), addedge((num<<1)|1, (i*i-num)<<1, 0); } flow += dinic(s, t); } cout << --num; for(int i = 1; i <= num; ++i) { if(!vis[i]) { cout << "\n"; for(int u = i; u&&u!=t>>1; u = pre[u]) { vis[u] = true; cout << u << " "; } } } } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); //cout.flush(); return 0; }
标签：head cli 覆盖 ace owb als include 多少 节点
原文地址：https://www.cnblogs.com/GRedComeT/p/12273689.html
