题目

Luogu
LOJ
Acwing

思路

代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int N = 1010, M = N * N, INF = 1e9;
int Cases, n, m, S, T, col[N], row[N], cnt;
int h[N], t[M], p[M], f[M], idx;
char g[N][N];
void add(int a, int b, int c) { f[idx] = c, t[idx] = b, p[idx] = h[a], h[a] = idx++; }
void add(int a, int b, int c, int d) { add(a, b, c), add(b, a, d); }
map<string, int> id;
set<string> same;
void Process(string str, int type, int x) {
	int n = str.size(), flag;
	for (int i = 0, r; i < n; i = r) {
		if (str[i] == '_') { r = i + 1; continue; }
		for (r = i; (r < n && str[r] != '_'); r++);
		string a = str.substr(i, r - i); 
		string b = a; reverse(b.begin(), b.end());
		(b < a) ? swap(a, b), flag = -1 : flag = 1;
		if (a == b) { same.insert(a); continue; }
		if (!id.count(a)) id[a] = ++cnt, id[b] = ++cnt, 
						  add(id[a], id[b], 2, 0);
		flag *= type;
		if (flag == 1) add(S, id[a], INF, 0);
		if (flag == -1) add(id[b], T, INF, 0);
		if (flag == 0) add(id[b], x, INF, 0), add(x, id[a], INF, 0);
	}
}
int d[N], cur[N], q[N];
bool BFS() {
	int hh = 0, tt = -1;
	memset(d, -1, sizeof d);
	d[S] = 0, q[++tt] = S, cur[S] = h[S];
	while (hh <= tt) {
		int u = q[hh++];
		for (int i = h[u], v; v = t[i], i != -1; i = p[i]) {
			if (d[v] != -1 || !f[i]) continue;
			d[v] = d[u] + 1, cur[v] = h[v], q[++tt] = v;
			if (v == T) return true;
		}
	}
	return false;
}
int DFS(int u, int limit) {
	if (u == T) return limit;
	int flow = 0;
	for (int i = cur[u], v; v = t[i], i != -1 && flow < limit; i = p[i]) {
		cur[u] = i;
		if (d[v] == d[u] + 1 && f[i]) {
			int t = DFS(v, min(f[i], limit - flow));	
			if (!t) d[v] = -1;
			f[i] -= t, f[i ^ 1] += t, flow += t;
		}
	}
	return flow;
}		
int Dinic() {	
	int flow = 0, res = 0;
	while (BFS()) while (flow = DFS(S, INF)) res += flow;
	return res;
}
int main() {
	cin >> Cases;
	while (Cases--) {
		memset(h, -1, sizeof h), idx = 0;
		same.clear(), id.clear();
		cin >> n >> m;
		S = 1, T = 2, cnt = 2 + n + m;
		for (int i = 1; i <= n; i++) cin >> row[i];
		for (int i = 1; i <= m; i++) cin >> col[i];
		for (int i = 1; i <= n; i++) cin >> g[i] + 1;
		string str;
		for (int i = 1; i <= n; i++) // row
			str = g[i] + 1, Process(str, row[i], i + 2);
		for (int i = 1; i <= m; i++) { // col
			str.clear();
			for (int j = 1; j <= n; j++)
				str += g[j][i];
			Process(str, col[i], i + n + 2);
		}
		cout << Dinic() + same.size() << endl;
	}
	return 0;
}

标签：SDOI,idx,int,flag,句子,add,str,2016,id
来源： https://www.cnblogs.com/Proteinlzl/p/15012567.html