【poj1804】 Brainman

http://poj.org/problem?id=1804 (題目鏈接)

題意：求逆序對

Solution1
　　歸併排序。
　　每次合併時計算逆序對。
　　
代碼：

// 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
　　樹狀數組。

代碼：

// 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;
}