hdu 5637 Transform 最短路

题目链接

异或的性质。 求s到t的最少步骤， 等价于求0到s^t的最少步骤。



通过最少的步骤达到s^t的状态， 等价于求0到s^t的最短路。 先将最短路求出来然后O(1)查询。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e6+5;
int vis[maxn], dis[maxn], a[16], head[maxn], num;
struct node
{
    int to, nextt;
}e[maxn*10];
void init() {
    mem1(head);
    num = 0;
}
void add(int u, int v) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void spfa() {
    queue<pll> q;
    q.push(mk(0, 0));
    mem(vis);
    mem2(dis);
    dis[0] = 0;
    vis[0] = 1;
    while(!q.empty()) {
        pll tmp = q.front(); q.pop();
        for(int i = head[tmp.se]; ~i; i = e[i].nextt) {
            int v = e[i].to;
            if(vis[v])
                continue;
            vis[v] = 1;
            if(dis[v]>dis[tmp.se]+1) {
                dis[v] = dis[tmp.se]+1;
                q.push(mk(-dis[v], v));
            }
        }
    }
}
void solve() {
    int n, m;
    cin>>n>>m;
    for(int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    for(int i = 0; i <= 1e5; i++) {
        for(int j = 0; j < n; j++) {
            add(i, a[j]^i);
        }
        for(int j = 0; j < 17; j++) {
            add(i, i^(1<<j));
        }
    }
    spfa();
    ll ans = 0;
    int s, t;
    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &s, &t);
        s ^= t;
        ans += i*dis[s];
        ans %= mod;
    }
    cout<<ans<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--) {
        init();
        solve();
    }
    return 0;
}