HDU 5627 Clarke and MST（贪心、连通性）

题意:

N,M≤3×105的无向图,求权值位与(&)运算的最大生成树

分析:

首先与运算的性质就是参与运算的该位全是1结果才是1

先把边按照该位是不是1分类,对于生成怎么样的生成树我们不关心,只要抓住如果这些边能让这个图连通,那么一定能生成树

所以从高位到低位枚举这些边是不是连通即可

时间复杂度为O(30mα(n))

代码:

//
//  Created by TaoSama on 2016-02-13
//  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 << endl
const int N = 3e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
struct Edge {
int u, v, c;
};
vector<Edge> G[30];

struct DSU {
int p
;
void init() {
for(int i = 1; i <= n; ++i) p[i] = i;
}
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;
p[x] = y;
}
int count() {
int cc = 0;
for(int i = 1; i <= n; ++i) cc += p[i] == i;
return cc;
}
} dsu;

bool isConnected(int x, int p) {
dsu.init();
for(Edge& e : G[x]) {
if(e.c >> p & 1) {
dsu.unite(e.u, e.v);
}
}
return dsu.count() == 1;
}

int solve(int x) {
int ret = 1 << x;
for(int i = x - 1; ~i; --i) {
if(isConnected(x, i))
ret |= 1 << i;
}
return ret;
}

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);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for(int i = 0; i < 30; ++i) G[i].clear();
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
for(int j = 0; j < 30; ++j)
if(c >> j & 1) G[j].push_back((Edge) {u, v, c});
}
int ans = 0;
for(int i = 29; ~i; --i) {
if(isConnected(i, i)) {
ans = solve(i);
break;
}
}
printf("%d\n", ans);
}
return 0;
}