Known Notation ZOJ - 3829 (后缀表达式,贪心)

大意:给定后缀表达式, 每次操作可以添加一个字符, 可以交换两个字符的位置, 相邻数字可以看做一个整体也可以分开看, 求合法所需最少操作数.

数字个数一定为星号个数+1, 添加星号一定不会更优.

先判断若星号过多, 直接在最左边添上数字, 遍历过程中若星号还多的话把星号与右侧数字交换.

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+10;

#else

const int N = 111;

#endif

int n;

char s[N];

void work() {

	scanf("%s", s+1);

	n = strlen(s+1);

	int star = 0, num = 0;

	REP(i,1,n) {

		if (s[i]=='*') ++star;

		else ++num;

	}

	int left_num = 0, ans = 0;

	if (num<=star) {

		left_num += star-num+1;

		ans += left_num;

	}

	int now = n;

	REP(i,1,n) {

		while (i<now&&s[now]=='*') --now;

		if (s[i]=='*') {

			if (--left_num<1) {

				++ans,--now;

				left_num+=2;

			}

		}

		else ++left_num;

	}

	printf("%d\n", ans);

}

int main() {

	int t;

	scanf("%d", &t);

	while (t--) work();

}