HDU 4619 Warm up 2（贪心、并查集 | 二分图最大独立集）

题意:

1∗2的多米诺骨牌,n≤1000个横向的,m≤1000个纵向的
横向的之间互相没有交点,纵向也是
现要删掉几个,求剩下的最都没有交点的个数

分析:

一开始想只删纵的或者横的,然后反例就是这种图形

显然答案是4,删掉一横一纵
然后新的贪心姿势就来了,还是由于题目那个已知
应该是每个连通块,只留下横的或者竖的,留下多的那个
然后就能搞了,时间复杂度是O(n+m)

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其实这是一个二分图,然后没交点就是二分图最大独立集。。。
时间复杂度是O(n2),边数最多是2n

贪心代码:

////  Created by TaoSama on 2016-02-28//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m;struct DSU {    int n, m;    int r[N], p[N];    int row[N], col[N];    void init(int _n, int _m) {        n = _n; m = _m;        for(int i = 1; i <= n + m; ++i) {            p[i] = i;            r[i] = 0;            if(i <= n) row[i] = 1, col[i] = 0;            else row[i] = 0, col[i] = 1;        }    }    int find(int x) {        return p[x] = p[x] == x ? x : find(p[x]);    }    void unite(int x, int y) {        x = find(x), y = find(y);        if(x == y) return;        if(r[x] > r[y]) swap(x, y);        p[x] = y;        if(r[x] == r[y]) ++ r[y];        row[y] += row[x]; row[x] = 0;        col[y] += col[x]; col[x] = 0;    }    int get() {        int ret = 0;        for(int i = 1; i <= n + m; ++i) {//          printf("%d: %d %d\n", i, row[i], col[i]);            ret += max(row[i], col[i]);        }        return ret;    }} dsu;int row[105][105];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &n, &m) == 2 && (n || m)) {        dsu.init(n, m);        memset(row, 0, sizeof row);        for(int i = 1; i <= n; ++i) {            int x, y; scanf("%d%d", &x, &y);            row[x][y] = i, row[x + 1][y] = i;        }        for(int i = 1; i <= m; ++i) {            int x, y; scanf("%d%d", &x, &y);            if(row[x][y + 1]) dsu.unite(n + i, row[x][y + 1]);            if(row[x][y]) dsu.unite(n + i, row[x][y]);        }        int ans = dsu.get();        printf("%d\n", ans);    }    return 0;}