uva 1608 Non-boring sequences
2017-07-30 08:35
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解题思路:
不无聊的序列,这是刘汝佳紫书上的一道题,首先先找出一个独一无二的数字,那么只要跨越了这个数字的序列就是不无聊的,这个时候,再去找这个数字左边的序列和这个数字右边的序列,看是否是无聊的
代码:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
const int MAX = 200010;
int N,T;
int a[MAX],l[MAX],r[MAX];
bool solve(int left, int right)
{
if(left >= right)
return true;
for(int i = 0; i <= (right - left) / 2; ++i){
if(l[left + i] < left && r[left + i] > right)
return solve(left, left + i - 1) && solve(left + i + 1,right);
if(l[right - i] < left && r[right - i] > right)
return solve(left,right - i - 1) && solve(right - i + 1, right);
}
return false;
}
int main(void)
{
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i = 0; i < N; ++i)
scanf("%d",&a[i]);
map<int, int> M;
for(int i = 0; i < N; ++i){
if(!M.count(a[i])) l[i] = -1;
else l[i] = M[a[i]];
M[a[i]] = i;
}
M.clear();
for(int i = N - 1; i >= 0; --i){
if(!M.count(a[i])) r[i] = N;
else r[i] = M[a[i]];
M[a[i]] = i;
}
if(solve(0,N-1))
puts("non-boring");
else
puts("boring");
}
return 0;
}
不无聊的序列,这是刘汝佳紫书上的一道题,首先先找出一个独一无二的数字,那么只要跨越了这个数字的序列就是不无聊的,这个时候,再去找这个数字左边的序列和这个数字右边的序列,看是否是无聊的
代码:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
const int MAX = 200010;
int N,T;
int a[MAX],l[MAX],r[MAX];
bool solve(int left, int right)
{
if(left >= right)
return true;
for(int i = 0; i <= (right - left) / 2; ++i){
if(l[left + i] < left && r[left + i] > right)
return solve(left, left + i - 1) && solve(left + i + 1,right);
if(l[right - i] < left && r[right - i] > right)
return solve(left,right - i - 1) && solve(right - i + 1, right);
}
return false;
}
int main(void)
{
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i = 0; i < N; ++i)
scanf("%d",&a[i]);
map<int, int> M;
for(int i = 0; i < N; ++i){
if(!M.count(a[i])) l[i] = -1;
else l[i] = M[a[i]];
M[a[i]] = i;
}
M.clear();
for(int i = N - 1; i >= 0; --i){
if(!M.count(a[i])) r[i] = N;
else r[i] = M[a[i]];
M[a[i]] = i;
}
if(solve(0,N-1))
puts("non-boring");
else
puts("boring");
}
return 0;
}
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