HDU 1394 Minimum Inversion Number 树状数组&&线段树

题目给了你一串序列，然后每次 把最后一个数提到最前面来，直到原来的第一个数到了最后一个，每次操作都会产生一个新的序列，这个序列具有一个逆序数的值，问最小的你逆序数的值为多少

逆序数么 最好想到的是树状数组，敲了一把很快，注意把握把最后一个数提上来对逆序数的影响即可，

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#include<cctype>#define ll long long#define LL __int64#define eps 1e-8#define inf 0xfffffff//const LL INF = 1LL<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;int n;int c[10000 + 5];int num[10000 + 5];void init() {memset(c,0,sizeof(c));memset(num,0,sizeof(num));}int lowbit(int x) {return x&(-x);}void add(int i,int val) {while(i <= n) {c[i] += val;i += lowbit(i);}}int get_sum(int i) {int sum = 0;while(i > 0) {sum += c[i];i -= lowbit(i);}return sum;}int main() {while(scanf("%d",&n) == 1) {init();int ans = 0;for(int i=1;i<=n;i++) {scanf("%d",&num[i]);num[i]++;add(num[i],1);ans += (i - get_sum(num[i]));}int minn = ans;for(int i=n;i>1;i--) {ans = ans + num[i] + num[i] - n - 1;minn = min(ans,minn);}printf("%d\n",minn);}return 0;}

线段树：

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#include<cctype>#define ll long long#define LL __int64#define eps 1e-8#define inf 0xfffffff//const LL INF = 1LL<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;const int N = 10000 + 5;int num[N];typedef struct Node {    int a;    int l,r;};Node tree[N * 4];void init() {    memset(tree,0,sizeof(tree));    memset(num,0,sizeof(num));}void cal(int id) {    tree[id].a = min(tree[id<<1].a,tree[id<<1|1].a);}void build(int l,int r,int id) {    tree[id].l = l;    tree[id].r = r;    tree[id].a = 0;    if(l == r) return ;    int mid = (l + r)/2;    build(l,mid,id<<1);    build(mid+1,r,id<<1|1);}void updata(int w,int id) {    if(tree[id]. l == w && tree[id].r == w) {        tree[id].a = 1;return;    }    int mid = (tree[id].l + tree[id].r)/2;    if(w <= mid) updata(w,id<<1);    else updata(w,id<<1|1);    tree[id].a = tree[id<<1].a + tree[id<<1|1].a;}int query(int l,int r,int id) {    if(l <= tree[id].l && r >= tree[id].r)return tree[id].a;    int mid = (tree[id].l + tree[id].r)/2;    int ans1 = 0,ans2 = 0;    if(l <= mid) ans1 = query(l,r,id<<1);    if(r > mid) ans2 = query(l,r,id<<1|1);    return ans1 + ans2;}int main() {    int n;    while(scanf("%d",&n) == 1) {        init();        build(1,n,1);        int ans = 0;        for(int i=0;i<n;i++) {            scanf("%d",&num[i]);            ans += query(num[i] + 1,n-1,1);            updata(num[i],1);        }        int minn = ans;        for(int i=0;i<n;i++) {            ans = ans + n - 2 * num[i] - 1;            minn = min(ans,minn);        }        printf("%d\n",minn);    }    return 0;}