[CodeForces 344D Alternating Current]栈

题意：两根导线绕在一起，问能不能拉成两条平行线，只能向两端拉不能绕

思路：从左至右，对+-号分别进行配对，遇到连续的两个“+”或连续的两个“-”即可消掉，最后如果全部能消掉则能拉成平行线。拿两根线绕一下就理解了，也可以一根拉成直线，另一根围着它绕，然后观察能拉成直线的条件。用栈实现就行。

 

 

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#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

char s[123456];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    while (~scanf("%s", s)) {
        stack<char> S;
        for (int i = 0; s[i]; i ++) {
            char ch = s[i];
            if (!S.empty() && ch == S.top()) S.pop();
            else S.push(ch);
        }
        puts(S.empty()? "Yes" : "No");
    }
    return 0;
}
posted @ 2015-08-12 19:01  jklongint  阅读(274)  评论(0)  编辑  收藏  举报