LianLianKan[HDU4272]

LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1760    Accepted Submission(s): 547

 

Problem Description
I like playing game with my friend, although sometimes looks pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.
Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with one same-value element. After pop them from stack, all left elements will fall down. Although the game seems to be interesting, it's really naive indeed.

 

                                    
To prove I am a wisdom among my friend, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 6 with it.
Before the game, I want to check whether I have a solution to pop all elements in the stack.
 

 

Input
There are multiple test cases.
The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)
The next line contains N integer ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)
 

 

Output
For each test case, output “1” if I can pop all elements; otherwise output “0”.
 

 

Sample Input
2
1 1
3
1 1 1
2
1000000 1
 

 

Sample Output
1
0
0
 

 

Source
2012 ACM/ICPC Asia Regional Changchun Online
 

 

Recommend
liuyiding

想了半天不知咋办,在网上搜了一下原来这样也可以...

#include <algorithm>     
#include <iostream>    
#include <cstring>   
#include <cstdio>   
#include <vector>   
using namespace std;  
#define MAXN 1010      
#define INF 0xFFFFFFF   
  
int n;  
vector<int>v;  
  
void solve() {  
    int flag, i;  
    vector<int>::iterator it1, it2;  
    while (1) {  
        flag = 0;  
        if (v.size() <= 1) break;/*只有一个元素的肯定不行*/  
        it1 = v.begin();  
        for (; it1 != v.end(); it1++) {  
            for (it2 = it1 + 1, i = 1; i < 6 && it2 != v.end(); i++ , it2++) {  
                if (*it1 == *it2) {  
                    v.erase(it1);/*这里删除了it1后it2会往回移动一个*/  
                    v.erase(it2-1);/*所以由上面的可知这里删除it2-1位置*/  
                    flag = 1 ; break;  
                }  
            }  
            if (flag) break;  
        }  
        if (!flag || !v.size()) break;/*如果flag为0或v为空退出*/  
    }  
    if (v.size()) printf("0
");  
    else printf("1
");  
}  
  
int main() {  
    //freopen("input.txt", "r", stdin);   
    int tmp;  
    while (scanf("%d", &n) != EOF) {  
        v.clear();  
        for (int i = 0; i < n; i++) {  
            scanf("%d", &tmp);  
            v.push_back(tmp);/*全部插入vector*/  
        }  
        solve();  
    }  
    return 0;  
}