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传送门:http://poj.org/problem?id=3041
题意:
坐标系上有k个点,坐标分别为(xi,yi),每次射击可以选择一列或者一行,问消灭需要多少次能把所有点都射完。
思路:
对于每个点,由行向列连一条边来构造一个二分图。根据题意只需要行或者列被包括即可,也就是对于一条边只要有一个顶点在点集合里。那么建完图后只需要求最小顶点覆盖就可以了。
之后再利用在二分图中最小顶点覆盖=最大匹配就可以了。
AC代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs id*2+1
#define Mod(a,b) a<b?a:a%b+b
using namespace std;
const ll M = 998244353;
const ll INF = 1e9 + 10;
const int N = 1010;
const double e = 10e-6;
int n, k, r, c;
bool used[N];
struct edge { int to; ll cap; int rev; };
vector<edge>G[N];
int level[N], iter[N];
void addEdge(int from, int to, ll cap)
{
G[from].push_back(edge{ to,cap,(int)G[to].size() });
G[to].push_back(edge{ from,0,(int)G[from].size() - 1 });
}
void init()
{
for (int i = 0; i < N; i++)
G[i].clear();
}
void bfs(int s)
{
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0; que.push(s);
while (!que.empty()) {
int v = que.front(); que.pop();
int len = G[v].size();
for (int i = 0; i < len; i++) {
edge &e = G[v][i];
if (e.cap > 0 && level[e.to] < 0) {
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}
ll dfs(int v, int t, ll f)
{
if (v == t) return f;
int len = G[v].size();
for (int &i = iter[v]; i < len; i++) {
edge &e = G[v][i];
if (e.cap > 0 && level[v] < level[e.to]) {
ll d = dfs(e.to, t, min(f, e.cap));
if (d > 0) {
e.cap -= d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
ll maxFlow(int s, int t)
{
ll flow = 0;
while (1) {
bfs(s);
if (level[t] < 0) return flow;
memset(iter, 0, sizeof(iter));
ll f;
while ((f = dfs(s, t, INF)) > 0)
flow += f;
}
}
int main()
{
while (~scanf("%d%d", &n, &k)) {
memset(used, false, sizeof(used));
init();
int S = 0, T = 1;
while (k--) {
scanf("%d%d", &r, &c);
addEdge(2 * r, 2 * c + 1, 1);
if (!used[2 * r]) {
addEdge(S, 2 * r, 1);
used[2 * r] = true;
}
if (!used[2 * c + 1]) {
addEdge(2 * c + 1, T, 1);
used[2 * c + 1] = true;
}
}
ll ans = maxFlow(S, T);
printf("%lld\n", ans);
}
return 0;
}