给定长度为n的01串s，有两种操作：1、交换相邻的两个字符，花费为1e12；2、删除一个字符，花费为1e12 + 1，求使s不递减的最少花费

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e12, maxm = 4e4 + 5, mod = 998244353, N = 1e6;
int a[505][505], b[maxn];
// bool vis[maxn];
int n, m;
string s;
int cnt[maxn][2];

void solve(){
    int res = 0;
    int k;
    int x;
    int q;
    // cin >> n;
    cin >> s;
	n = s.size();
	s = " " + s;
	for(int i = 1; i <= n; i++){
		cnt[i][0] = cnt[i - 1][0];
		if(s[i] == '0'){
			cnt[i][0]++;
		}
	}
	cnt[n + 1][1] = 0;
	for(int i = n; i >= 1; i--){
		cnt[i][1] = cnt[i + 1][1];
		if(s[i] == '1'){
			cnt[i][1]++;
		}
	}
	res = (n - cnt[1][1]) * (inf + 1);
	for(int i = 1; i <= n; i++){
		int tmp = 1e18;
		if(s[i] == '0'){
			tmp = (n - cnt[i][0] - cnt[i + 1][1]) * (inf + 1);
		}
		else{
			if(i + 1 <= n && s[i + 1] == '0'){
				tmp = inf + (n - cnt[i][0] - cnt[i + 1][1] - 2) * (inf + 1);
			}
		}
		res = min(res, tmp);
	}
	cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    // fac[0] = 1;
    // for(int i = 1; i <= N; i++){
    //     fac[i] = fac[i - 1] * i % mod;
    // }
    // inv[N] = qpow(fac[N], mod - 2);
    // for(int i = N - 1; i >= 0; i--){
    //     inv[i] = inv[i + 1] * (i + 1) % mod;
    // }
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}