ZJU 3829 - Known Notation (贪心)
题意
输出最少的操作次数,使得成为逆波兰表达式
思路
显然只有缺少数字的时候才用到插入,不然就用交换。
交换的时候从最后面的数字开始交换。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 2000 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int fun[MAXN];
char str[MAXN];
int main()
{
//ROP;
ios::sync_with_stdio(0);
int T, i, j;
cin >> T;
while (T--)
{
string str;
cin >> str;
int num = 0, ope = 0, pos = 0;
for (i = SZ(str) - 1; i >= 0; i--)
{
if (str[i] != '*')
{
fun[pos++] = i;
num++;
}
else ope++;
}
if (num == SZ(str))
{
cout << "0" << endl;
continue;
}
int ans = max(0, ope + 1 - num);
int curNum = 0, curOp = 0;
curNum = ans;
pos = 0;
for (i = 0; i < SZ(str); i++)
{
if (str[i] != '*') curNum++;
else
{
if (curOp + 1 >= curNum)
{
swap(str[i], str[fun[pos++]]);
ans++;
curNum++;
}
else curOp++;
}
}
if (str[SZ(str) - 1] != '*') ans++;
cout << ans << endl;
}
return 0;
}