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Hdu1394Minimum Inversion Number线段树

  这个网上一搜一大堆,就是先求一个,其余的for一遍搞出来。

#include<stdio.h>#include<stdlib.h>#define max 5555int sum[max * 4];int min(int a, int b){    if (a>b) return b;    else return a;}void fuqin(int a){    sum[a] = sum[a * 2] + sum[a * 2 + 1];}void build(int l, int r, int a){    void fuqin(int);    sum[a] == 0;    if (l == r)        return;    int mid = (l + r) / 2;    build(l, mid, a * 2);    build(mid + 1, r, a * 2 + 1);    fuqin(a);}void up(int q, int l, int r, int a){    void fuqin(int);    if (l == r)    {        sum[a]++;        return;    }    int mid = (l + r) / 2;    if (q <= mid) up(q, l, mid, a * 2);    else up(q, mid + 1, r, a * 2 + 1);    fuqin(a);}int ask(int L, int R, int l, int r, int a){    if (L <= l&&r <= R)        return sum[a];    int mid = (l + r) / 2;    int ans = 0;    if (mid >= L) ans += ask(L, R, l, mid, a * 2);    if (mid<R) ans += ask(L, R, mid + 1, r, a * 2 + 1);    return ans;}void main(){    int n, i, x[5555], ans = 0, q;    int min(int, int);    void build(int, int, int);    void tianjia(int, int, int, int);    int ask(int, int, int, int, int);    while (scanf("%d", &n) != EOF)    {        ans = 0;        memset(sum, 0, sizeof(sum));        build(0, n - 1, 1);        for (i = 0; i<n; i++)        {            scanf("%d", &x[i]);            ans += ask(x[i], n - 1, 0, n - 1, 1);            up(x[i], 0, n - 1, 1);        }        q = ans;        for (i = 0; i<n; i++)        {            q = q + (n - 1 - x[i]) - x[i];            ans = min(q, ans);        }        printf("%d\n", ans);    }}