GYM 101669F - Binary Transformations

做法：如果不存在一个位置p \((a[p]=1,b[p]=1)\)，那么答案就是贪心的先把所有的1，按价值从大到小变为0,所有的0,按价值从小到大变为1。如果存在一些位置p，我们就枚举一开始把多少p转成0,显然价值越大的p越优。现在考虑如何模拟，我们可以用2个set，一个维护一开始要从0变1的数，另一个维护最后要由1变0的数，插入\(O(log n)\)，遍历\(O(n)\)，总的复杂度\(O(n^2)\)

#include <bits/stdc++.h>
#define pb push_back
typedef long long ll;
const int N = 5010;
using namespace std;
int n;
ll c[N], ans, sum;
char a[N], b[N];
bool cmp(ll a,ll b) {return a > b;}
vector<ll> v3;
multiset<ll> tmp1,tmp2;
ll cal(int x) {ll ans = 0, ss = sum;if(x) tmp1.insert(v3[x-1]),tmp2.insert(v3[x-1]);multiset<ll>::iterator it1;multiset<ll>::reverse_iterator it2;for(it2 = tmp1.rbegin(); it2 != tmp1.rend(); ++it2) {ss -= (*it2); ans += ss;}for(it1 = tmp2.begin(); it1 != tmp2.end(); ++it1) {ss += (*it1); ans += ss;}return ans;
}
int main() {scanf("%d",&n);for(int i = 1; i <= n; ++i) scanf("%lld",&c[i]);scanf(" %s",a+1);scanf(" %s",b+1);for(int i = 1; i <= n; ++i) {if(a[i] != b[i]) {if(a[i] == '1' && b[i] == '0') tmp1.insert(c[i]);else tmp2.insert(c[i]);}else if(a[i] == '1' && b[i] == '1') v3.pb(c[i]);if(a[i] == '1') sum+=c[i];}sort(v3.begin(),v3.end(),cmp);ans = (ll)(1e18);int v3n = v3.size();for(int i = 0; i <= v3n; ++i) ans = min(ans,cal(i));printf("%lld\n",ans);return 0;
}

转载于:https://www.cnblogs.com/RRRR-wys/p/9720412.html

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GYM 101669F - Binary Transformations

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