UVaLive 7637 Balanced String (构造)

题意：给定一个括号的序列，原先的序列是碰到左括号加1，碰到右括号减1，然后把序列打乱，让你找出字典序最小的一个答案。

析：直接从第一个括号判断就好了，优先判断左括号，如果不行就加右括号。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
map<int, int> mp;

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    mp.clear();
    for(int i = 0; i < n; ++i)  scanf("%d", a+i), ++mp[a[i]];
    printf("Case %d: ", kase);
    if((n&1) || !mp.count(1)){  printf("invalid\n");  continue;   }
    bool ok = true;
    --mp[1];
    int pos = 1;
    for(int i = 1; i < n && ok; ++i){
      if(mp.count(pos+1) && mp[pos+1])  ++pos, --mp[pos];
      else if(mp.count(pos-1) && mp[pos-1])  --pos, --mp[pos];
      else ok = false;
    }
    if(!ok && pos){  printf("invalid\n");  continue; }
    for(int i = 0; i < n; ++i)  ++mp[a[i]];
    printf("(");
    pos = 1;
    for(int i = 1; i < n; ++i){
      if(mp.count(pos+1) && mp[pos+1])  ++pos, --mp[pos], printf("(");
      else if(mp.count(pos-1) && mp[pos-1])  --pos, --mp[pos], printf(")");
    }
    printf("\n");
  }
  return 0;
}