HDU 5637 Transform(bfs)

题意:

$N\le 15个整数A_i \le 10^5,对于一个数x,2种操作:$
$1.翻转二进制位中的1个位$
$2.x\oplus A_i,1次选择1个A_i,\oplus为二进制异或$
$Q\le 10^5询问,s\to t的最小操作数$

分析:

$对于第一种操作其实就是x\oplus 2^i,i\in [0,17)$
$多次询问其实我们可以bfs预处理,d[x]为到x的最小步数$
$然后我们发现ans(s,t)=d[s\oplus t]$

代码:

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//
// Created by TaoSama on 2016-04-06
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#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 = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m, a[16];
int d[1 << 17];
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 = 1; i <= n; ++i) scanf("%d", a + i);
memset(d, 0x3f, sizeof d);
queue<int> q; q.push(0);
d[0] = 0;
while(q.size()) {
int u = q.front(); q.pop();
for(int i = 0; i < 17; ++i) {
int v = u ^ (1 << i);
if(d[v] == INF) {
d[v] = d[u] + 1;
q.push(v);
}
}
for(int i = 1; i <= n; ++i) {
int v = u ^ a[i];
if(d[v] == INF) {
d[v] = d[u] + 1;
q.push(v);
}
}
}
int ans = 0;
for(int i = 1; i <= m; ++i) {
int s, t; scanf("%d%d", &s, &t);
ans += 1LL * i * d[s ^ t] % MOD;
if(ans >= MOD) ans -= MOD;
}
printf("%d\n", ans);
}
return 0;
}


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