一个很眼熟的贪心 首先一个序列对他编好号后，末状态编号的逆序对数就是交换相邻2个达到这个状态需要的最少次数 将A序列1~n编号，再对B序列按顺序每个字母赋这个字母有的所有编号里最小的这样能使得逆序对数最少，然后求B的逆序对数
code：
#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define lowbit(x) x&(-x)using namespace std;const int maxn = 1100000;queue<int>Q[26];int n;int s[maxn];struct node{ int x,i; node(){} node(int _x,int _i){x=_x;i=_i;}}a[maxn];bool cmp(node x,node y){return x.x<y.x;}void add(int x){for(;x<=n;x+=lowbit(x))s[x]++;}int query(int x){ int r=0; for(;x;x-=lowbit(x)) r+=s[x]; return r;}int main(){ char str; scanf("%d",&n); getchar(); for(int i=1;i<=n;i++) { scanf("%c",&str); Q[str-'A'].push(i); }getchar(); for(int i=1;i<=n;i++) { scanf("%c",&str); int j=str-'A'; int x=Q[j].front(); Q[j].pop(); a[i]=node(x,i); } sort(a+1,a+n+1,cmp); ll re=0; for(int i=n;i>=1;i--) { re+=(ll)query(a[i].i); add(a[i].i); } PRintf("%lld/n",re); return 0;}新闻热点
疑难解答
