标签:style http io color os ar for sp on
题目链接:LightOJ 1269 - Consecutive Sum
题目大意:给定一个序列,选定一段区间的亦或和,输出最大和最小。
解题思路:最大很简单,对所有前缀建立字典树,然后尽量往反向走;最小则需要往正向走,并且向正向走的时候要扣
除自己本身。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 50005 * 32;
const int sigma_size = 2;
struct Tire {
int sz;
int g[maxn][sigma_size];
int c[maxn];
void init();
void insert(int s);
int findMax(int s);
int findMin(int s);
}T;
int N, A[maxn];
int main () {
int cas, x;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
T.init();
scanf("%d", &N);
for (int i = 1; i <= N; i++) {
scanf("%d", &x);
A[i] = A[i-1] ^ x;
}
for (int i = 0; i <= N; i++)
T.insert(A[i]);
int ansMax = 0, ansMin = (1<<31)-1;
for (int i = 0; i <= N; i++) {
ansMax = max(ansMax, T.findMax(A[i]));
ansMin = min(ansMin, T.findMin(A[i]));
}
printf("Case %d: %d %d\n", kcas, ansMax, ansMin);
}
return 0;
}
void Tire::init() {
sz = 1;
c[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Tire::findMin(int s) {
int ret = 0, u = 0;
for (int i = 30; i >= 0; i--) {
int v = (s>>i) & 1;
if (g[u][v] == 0 || (g[u][v^1] && c[g[u][v]] < 2)) {
v = v ^ 1;
ret |= (1<<i);
}
u = g[u][v];
}
return ret;
}
int Tire::findMax(int s) {
int ret = 0, u = 0;
for (int i = 30; i >= 0; i--) {
int v = ((s>>i)&1)^1;
if (g[u][v])
ret |= (1<<i);
else
v = v ^ 1;
u = g[u][v];
}
return ret;
}
void Tire::insert(int s) {
int u = 0;
for (int i = 30; i >= 0; i--) {
int v = (s>>i) & 1;
if (g[u][v] == 0) {
c[sz] = 0;
memset(g[sz], 0, sizeof(g[sz]));
g[u][v] = sz++;
}
u = g[u][v];
c[u]++;
}
}
LightOJ 1269 - Consecutive Sum(字典树)
标签:style http io color os ar for sp on
原文地址:http://blog.csdn.net/keshuai19940722/article/details/40633295