题意:n个插座,m个设备及其插头类型,k种转换器,没有转换器的情况下插头只能插到类型名称相同的插座中,问最少剩几个不匹配的设备
lrj紫书里面讲得挺好的。
先跑一遍floyd,看看插头类型a能否转换为b
然后构造网络
源点为0, 汇点为n + m + 1,源点连插头 容量为1,插座连汇点,容量为1
插头和插座能互相转换的容量为INF,跑一遍最大流 m - 最大流就是答案
顺便粘一下dinic的板子
#include <bits/stdc++.h> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < ‘0‘ || ch > ‘9‘) { if (ch == ‘-‘) f = -1; ch = getchar();} while (ch >= ‘0‘ && ch <= ‘9‘) { x = x * 10 + ch - 48; ch = getchar();} return x * f; } const int N = 500; const int INF = 0x3f3f3f3f; int n, m, k, tol, cnt; map<string, int> mp; vector<int> a; vector<int> b; bool f[N][N]; struct Edge { int v, f, next; } edge[N * 2]; int head[N], level[N], iter[N]; inline void init() { mp.clear(); a.clear(); b.clear(); tol = cnt = 0; memset(f, 0, sizeof(f)); memset(head, -1, sizeof(head)); } inline void addedge(int u, int v, int f) { edge[cnt].v = v; edge[cnt].next = head[u]; edge[cnt].f = f; head[u] = cnt++; } void floyd() { for (int k = 1; k <= tol; k++) { for (int i = 1; i <= tol; i++) { for (int j = 1; j <= tol; j++) { f[i][j] = f[i][j] || (f[i][k] && f[k][j]); } } } } bool bfs(int s, int t) { for (int i = 0; i <= t; i++) iter[i] = head[i], level[i] = -1; level[s] = 0; queue<int> que; que.push(s); while (!que.empty()) { int u = que.front(); que.pop(); for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].v, f = edge[i].f; if (f > 0 && level[v] == -1) { level[v] = level[u] + 1; que.push(v); } } } return level[t] != -1; } int dfs(int u, int t, int f) { if (!f || u == t) return f; int flow = 0, w; for (int i = iter[u]; ~i; i = edge[i].next) { iter[u] = i; int v = edge[i].v; if (level[v] == level[u] + 1 && edge[i].f > 0) { w = dfs(v, t, min(f, edge[i].f)); if (w == 0) continue; f -= w; edge[i].f -= w; edge[i^1].f += w; flow += w; if (f <= 0) break; } } return flow; } int dinic(int s, int t) { int ans = 0; while (bfs(s, t)) ans += dfs(s, t, INF); return ans; } int main() { int T = read(); while (T--) { init(); n = read(); for (int i = 1; i <= n; i++) { string s; cin >> s; mp[s] = i; a.emplace_back(i); tol++; } m = read(); for (int i = 1; i <= m; i++) { string str1, s; cin >> str1 >> s; if (!mp.count(s)) { mp[s] = ++tol; } b.emplace_back(mp[s]); } k = read(); for (int i = 1; i <= k; i++) { string s1, s2; cin >> s1 >> s2; if (!mp.count(s1)) mp[s1] = ++tol; if (!mp.count(s2)) mp[s2] = ++tol; f[mp[s1]][mp[s2]] = 1; } for (int i = 1; i <= tol; i++) f[i][i] = 1; floyd(); for (int i = 1; i <= m; i++) { addedge(0, i, 1); addedge(i, 0, 0); } for (int i = 1; i <= n; i++) { addedge(m + i, n + m + 1, 1); addedge(n + m + 1, m + i, 0); } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (!f[b[i]][a[j]]) continue; addedge(i + 1, m + j + 1, INF); addedge(m + j + 1, i + 1, 0); } } printf("%d\n", m - dinic(0, n + m + 1)); if (T) puts(""); } return 0; }
原文地址:https://www.cnblogs.com/Mrzdtz220/p/10799423.html
时间: 2024-11-06 07:34:18