【poj1804】 Brainman

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http://poj.org/problem?id=1804 (题目链接)

题意：求逆序对

Solution1
　　归并排序。
　　每次合并时计算逆序对。
　　
代码：

// poj1804#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#define LL long long#define inf 2147483640#define Pi acos(-1.0)#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);using namespace std;inline LL getint() {    int f,x=0;char ch=getchar();    while (ch<='0' || ch>'9') {if (ch=='-') f=-1;else f=1;ch=getchar();}    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}    return x*f;}const int maxn=1010;int a[maxn],tmp[maxn],ans,n;void solve(int l,int r) {    if (l<r) {        int mid=(l+r)>>1;        solve(l,mid);        solve(mid+1,r);        int i=l,j=mid+1,k=l;        while (i<=mid && j<=r) {            if (a[i]>a[j]) {tmp[k++]=a[j++];ans+=mid-i+1;}            else tmp[k++]=a[i++];        }        while (i<=mid) tmp[k++]=a[i++];        while (j<=r) tmp[k++]=a[j++];        for (int i=l;i<=r;i++) a[i]=tmp[i];    }}int main() {    int T,tt=0;scanf("%d",&T);    while (T--) {        scanf("%d",&n);        for (int i=1;i<=n;i++) scanf("%d",&a[i]);        ans=0;        solve(1,n);        printf("Scenario #%d:\n%d\n\n",++tt,ans);    }    return 0;}

Solution2
　　树状数组。

代码：

// poj1804#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#define LL long long#define inf 2147483640#define MOD 998244353#define Pi acos(-1.0)#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);using namespace std;inline LL getint() {    int f,x=0;char ch=getchar();    while (ch<='0' || ch>'9') {if (ch=='-') f=-1;else f=1;ch=getchar();}    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}    return x*f;}const int maxn=1010;int a[maxn],b[maxn],c[maxn],n,m;int lowbit(int x) {return x&-x;}void add(int x) {    for (int i=x;i<=m;i+=lowbit(i)) c[i]++;}int query(int x) {    LL res=0;    for (int i=x;i;i-=lowbit(i)) res+=c[i];    return res;}int main() {    int T,tt=0;scanf("%d",&T);    while (T--) {        scanf("%d",&n);        for (int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];        sort(b+1,b+1+n);        m=unique(b+1,b+1+n)-b-1;        for (int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+1+m,a[i])-b;        for (int i=1;i<=m;i++) c[i]=0;        LL ans=0;        for (int i=n;i>=1;i--) {            add(a[i]);            ans+=query(a[i]-1);        }        printf("Scenario #%d:\n%lld\n\n",++tt,ans);    }    return 0;}

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