hdu 4911 求逆序对数+树状数组

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http://acm.hdu.edu.cn/showproblem.php?pid=4911

给定一个序列，有k次机会交换相邻两个位置的数，问说最后序列的逆序对数最少为多少。

实际上每交换一次能且只能减少一个逆序对，所以问题转换成如何求逆序对数。

归并排序或者树状数组都可搞

树状数组：

先按大小排序后分别标号，然后就变成了求1~n的序列的逆序数，每个分别查询出比他小的用i减，在把他的值插入即可

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <vector>#include <iostream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define clr0(x) memset(x,0,sizeof(x))typedef long long LL;typedef pair<int,int> p;const int maxn=100005;LL f[maxn];int n;void add(int x,LL y){    for(;x<=n;x += x&(-x)) f[x]+=y;}LL sum(int x){    LL s=0;    for (;x;x -= x&(-x)) s+=f[x];    return s;}p a[maxn];int k;bool cmp(p i,p j){  return i.second < j.second;}int main(){  int i;  LL s;  while (~RD2(n,k)){    for (i=0;i<n;++i){        RD(a[i].first);        a[i].second=i;    }    sort(a,a+n);    for (i=0;i<n;++i)        a[i].first = i+1;    sort(a,a+n,cmp);    clr0(f);    for (s=i=0;i<n;++i){        s += i - sum(a[i].first);        add(a[i].first,1);    }    printf("%I64d\n",max(0LL,s-k));  }  return 0;}

归并排序：

每次归并发现需要前后交换时都给总的ret加上mid - mvl + 1,因为mvl到mid直接的数都比mvr下标上的数大

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <vector>#include <iostream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))typedef long long LL;const int maxn = 1e5+5;LL k;int n, a[maxn], b[maxn];LL merge_sort(int l, int r){    if (l == r)        return 0;    int mid = (l + r) / 2;    LL ret = merge_sort(l, mid) + merge_sort(mid+1, r);    int mvl = l, mvr = mid+1, mv = l;    while (mvl <= mid || mvr <= r) {        if (mvr > r || (mvl <= mid && a[mvl] <= a[mvr])) {            b[mv++] = a[mvl++];        } else {            ret += mid - mvl + 1;            b[mv++] = a[mvr++];        }    }    for (int i = l; i <= r; i++)        a[i] = b[i];    return ret;}int main () {    while (scanf("%d%I64d", &n, &k) == 2) {        for (int i = 1; i <= n; i++)            RD(a[i]);        printf("%I64d\n", max(merge_sort(1, n) - k, 0LL));    }    return 0;}

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