You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.

87 lines
2.5 KiB

2 years ago
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
#define x first
#define y second
const int N = 1010, M = N * N;
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int st[N], match[N];
int dfs(int u) {
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (st[v] == 1) continue;
st[v] = 1;
if (match[v] == -1 || dfs(match[v])) {
match[v] = u;
return 1;
}
}
return 0;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("POJ1548.in", "r", stdin);
#endif
int a, b;
while (true) {
memset(match, -1, sizeof match);
memset(st, 0, sizeof st);
memset(h, -1, sizeof h);
idx = 0;
vector<PII> vec;
while (true) {
cin >> a >> b;
if (a == 0 && b == 0) break;
if (a == -1 && b == -1) exit(0);
vec.push_back(make_pair(a, b));
}
sort(vec.begin(), vec.end());
vector<int> alls; // 存储所有待离散化的值
for (int i = 0; i < vec.size(); i++)
for (int j = i + 1; j < vec.size(); j++)
if (vec[j].y >= vec[i].y) alls.push_back(i), alls.push_back(j);
// 将所有值排序
sort(alls.begin(), alls.end());
// 去掉重复元素
alls.erase(unique(alls.begin(), alls.end()), alls.end());
// 重新捋着建边通过二分查找找出新的序号这样就可以使用i<=alls.size()了!
for (int i = 0; i < vec.size(); i++)
for (int j = i + 1; j < vec.size(); j++)
if (vec[j].y >= vec[i].y) {
int x = lower_bound(alls.begin(), alls.end(), i) - alls.begin();
int y = lower_bound(alls.begin(), alls.end(), j) - alls.begin();
add(x, y);
}
int res = 0;
for (int i = 0; i < alls.size(); i++) {
// 最开始黄海SB的以为二分后这里可以使用i<idx做为终止条件其实是SB到家了idx是边数是边数是边数
// 不是点数,不是点数,不是点数!
memset(st, 0, sizeof st);
if (dfs(i)) res++;
}
printf("%d\n", vec.size() - res);
}
return 0;
}