蓝桥集训之火柴排队
蓝桥集训之火柴排队
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核心思想：离散化+归并排序
- 由于数据范围较小10w 需要控制时间复杂度到nlogn 同时排两个数组会超时
- 所以将a数组离散化成顺序数组 b数组离散化后再归并排序求逆序对数量
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#include<iostream>#include <algorithm>#include <cstring>using namespace std;const int N = 100010 , MOD = 99999997;int n;int a[N],b[N],p[N],c[N];int find(int x) //离散化二分{int l=1,r=n;while(l<r){int mid = l+r >>1;if(p[mid] >=x ) r= mid;else l = mid + 1;}return l;}void work(int a[]) //离散化{for(int i=1;i<=n;i++) p[i] = a[i]; //将原本的a数组存下 仅用于排序sort(p+1,p+n+1);for(int i=1;i<=n;i++) a[i] = find(a[i]); //更新a数组元素为应该在的下标(根据元素大小)}int merge_sort(int l,int r) //求逆序对数量{if(l==r) return 0;int mid = l+r >>1;int res = (merge_sort(l,mid) + merge_sort(mid+1,r));int i = l,j = mid +1,k=0;while(i<=mid && j<=r){if(b[i] <= b[j]) p[k++] = b[i++];else p[k++] = b[j++] , res = (res + mid - i + 1) % MOD;}while(i<=mid) p[k++] = b[i++];while(j<=r) p[k++] = b[j++];for(int i=l,j=0;i<=r;i++,j++) b[i] = p[j];return res;}int main(){cin>>n;for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);work(a),work(b);for(int i=1;i<=n;i++) c[a[i]] = i; //c数组用来保存 对应关系for(int i=1;i<=n;i++) b[i] = c[b[i]]; //b数组按照 对应关系 更新cout<<merge_sort(1,n); //b数组归并return 0;}