线性基就解决子集异或问题，没了，可以最大最小，k大k小。
就是你插入不超过 \(\log a_i\) 个数大概能得到最大值。
假设你事先插入过 \(p_{i_1},p_{i_2}……,p_{i_k}\)
那么你插入一个 \(j , (j \leq i\{\})\)
实质上插入的是 \(x\oplus p_{i_1} \oplus p_{i_2}……\)
易证 \(2^i \leq p_i \leq 2^{(i+1)}\)
所以可以倒过来循环
至于怎么证明不知道，反正这样求最值是对的。
代码长这个样子
P3812 【模板】线性基
// by Isaunoya
#include
using namespace std;
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
#define int long long
const int _ = 1 << 21;
struct I {
char fin[_], *p1 = fin, *p2 = fin;
inline char gc() {
return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
}
inline I& operator>>(int& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c & 15);
while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
x = sign ? x : -x;
return *this;
}
inline I& operator>>(double& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c - 48);
while ((c = gc()) > 47) x = x * 10 + (c - 48);
if (c == '.') {
double d = 1.0;
while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
}
x = sign ? x : -x;
return *this;
}
inline I& operator>>(char& x) {
do
x = gc();
while (isspace(x));
return *this;
}
inline I& operator>>(string& s) {
s = "";
char c = gc();
while (isspace(c)) c = gc();
while (!isspace(c) && c != EOF) s += c, c = gc();
return *this;
}
} in;
struct O {
char st[100], fout[_];
signed stk = 0, top = 0;
inline void flush() { fwrite(fout, 1, top, stdout), fflush(stdout), top = 0; }
inline O& operator<<(int x) {
if (top > (1 << 20)) flush();
if (x < 0) fout[top++] = 45, x = -x;
do
st[++stk] = x % 10 ^ 48, x /= 10;
while (x);
while (stk) fout[top++] = st[stk--];
return *this;
}
inline O& operator<<(char x) {
fout[top++] = x;
return *this;
}
inline O& operator<<(string s) {
if (top > (1 << 20)) flush();
for (char x : s) fout[top++] = x;
return *this;
}
} out;
#define pb emplace_back
#define fir first
#define sec second
int n, a[64], p[64];
signed main() {
#ifdef _WIN64
freopen("testdata.in", "r", stdin);
#endif
in >> n;
rep(i, 1, n) in >> a[i];
rep(i, 1, n) {
for (int j = 50; ~j; --j) {
if ((a[i] & (1ll << j))) {
if (!p[j]) p[j] = a[i];
a[i] ^= p[j];
}
}
}
int ans = 0;
for (int i = 50; ~i; --i) ans = max(ans, ans ^ p[i]);
out << ans << '\n';
return out.flush(), 0;
} P4151 [WC2011]最大XOR和路径
找个出所有简单环，这样就可以和原有路径抵消掉
最后求异或，没了
// by Isaunoya
#include
using namespace std;
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
#define int long long
const int _ = 1 << 21;
struct I {
char fin[_], *p1 = fin, *p2 = fin;
inline char gc() {
return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
}
inline I& operator>>(int& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c & 15);
while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
x = sign ? x : -x;
return *this;
}
inline I& operator>>(double& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c - 48);
while ((c = gc()) > 47) x = x * 10 + (c - 48);
if (c == '.') {
double d = 1.0;
while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
}
x = sign ? x : -x;
return *this;
}
inline I& operator>>(char& x) {
do
x = gc();
while (isspace(x));
return *this;
}
inline I& operator>>(string& s) {
s = "";
char c = gc();
while (isspace(c)) c = gc();
while (!isspace(c) && c != EOF) s += c, c = gc();
return *this;
}
} in;
struct O {
char st[100], fout[_];
signed stk = 0, top = 0;
inline void flush() { fwrite(fout, 1, top, stdout), fflush(stdout), top = 0; }
inline O& operator<<(int x) {
if (top > (1 << 20)) flush();
if (x < 0) fout[top++] = 45, x = -x;
do
st[++stk] = x % 10 ^ 48, x /= 10;
while (x);
while (stk) fout[top++] = st[stk--];
return *this;
}
inline O& operator<<(char x) {
fout[top++] = x;
return *this;
}
inline O& operator<<(string s) {
if (top > (1 << 20)) flush();
for (char x : s) fout[top++] = x;
return *this;
}
} out;
#define pb emplace_back
#define fir first
#define sec second
template
inline void cmax(T& x, const T& y) {
(x < y) && (x = y);
}
template
inline void cmin(T& x, const T& y) {
(x > y) && (x = y);
}
int n, m;
const int N = 5e4 + 10;
const int M = 1e5 + 10;
int cnt = 0, head[N];
struct Edge {
int v, nxt, w;
} e[M << 1];
int d[N], p[65];
bool vis[N];
void add(int u, int v, int w) {
e[++cnt] = { v, head[u], w }, head[u] = cnt;
e[++cnt] = { u, head[v], w }, head[v] = cnt;
}
void insert(int val) {
for (int i = 63; ~i; --i)
if (val & (1ll << i)) {
if (!p[i]) p[i] = val;
val ^= p[i];
}
}
void dfs(int u, int cur) {
d[u] = cur, vis[u] = 1;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (vis[v])
insert(cur ^ e[i].w ^ d[v]);
else
dfs(v, cur ^ e[i].w);
}
}
int query(int val) {
int res = val;
for (int i = 63; ~i; --i) cmax(res, res ^ p[i]);
return res;
}
signed main() {
#ifdef _WIN64
freopen("testdata.in", "r", stdin);
#endif
in >> n >> m;
int u, v, w;
while (m--) {
in >> u >> v >> w, add(u, v, w);
}
dfs(1, 0);
out << query(d[n]) << '\n';
return out.flush(), 0;
} 留一些坑。
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