【POJ】1990-MooFest（树状数组or线段树）

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树状数组和线段树的解法比较，感觉不论在内存还是时间上都是树状数组占优

大题思路对 v从小到大进行排序，之后找之前的（也就是比v小的坐标）

v * sum（abs（xi - x））这样的话 abs无法处理，我们用另外一个树状数组记录在x ~ y区间牛的个数，之前那个记录在x ~ y区间内牛的坐标和

Accepted

48479

C++

1131

树状数组代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int maxn = 22222;int C1[maxn],C2[maxn];int n = 20005,nn;struct Cow{    int v,x;    friend bool operator < (Cow p,Cow q){        return p.v < q.v;    }}cow[maxn];int lowbit(int x){    return x & -x;}int sum(int C[],int x){    int ret = 0;    while(x > 0){        ret += C[x];        x -= lowbit(x);    }    return ret;}void add(int C[],int x,int d){    while(x <= n){        C[x] += d;        x += lowbit(x);    }}int main(){    scanf("%d",&nn);    memset(C1,0,sizeof(C1));    memset(C2,0,sizeof(C2));    for(int i = 0; i < nn; i++)        scanf("%d%d",&cow[i].v,&cow[i].x);    sort(cow,cow + nn);    LL ans = 0;    for(int i = 0; i < nn; i++){        int x = cow[i].x,v = cow[i].v;        int m1 = sum(C1,x),n1 = sum(C2,x);        int m2 = sum(C1,n) - m1,n2 = sum(C2,n) - n1;        ans += (((LL)n1 * x - m1) + ((LL)m2 - n2 * x)) * v;        add(C1,x,x);        add(C2,x,1);    }    printf("%I64d\n",ans);    return 0;}

Accepted

932110

C++

1723

线段树代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 20005;typedef long long LL;/*线段树解法*/#define lson (pos<<1)#define rson (pos<<1|1)int v1[maxn << 2],v2[maxn << 2];struct Node{    int x,v;    friend bool operator < (Node p,Node q){        return p.v < q.v;    }}node[maxn];void build(){    memset(v1,0,sizeof(v1));    memset(v2,0,sizeof(v2));}void pushup(int v[],int pos){    v[pos] = v[lson] + v[rson];}void update(int v[],int L,int R,int aim,int pos,int value){    if(L == R){        v[pos] += value;        return;    }    int mid = (L + R) >> 1;    if(aim <= mid)        update(v,L,mid,aim,lson,value);    else        update(v,mid + 1,R,aim,rson,value);    pushup(v,pos);}int query(int v[],int L,int R,int l,int r,int pos){    if(l <= L && R <= r){        return v[pos];    }    int mid = (L + R) >> 1;    int ans = 0;    if(l <= mid)        ans += query(v,L,mid,l,r,lson);    if(r  > mid)        ans += query(v,mid + 1,R,l,r,rson);    return ans;}int main(){    build();    int n,max_size = 0;    scanf("%d",&n);    for(int i = 0; i < n; i++){        scanf("%d%d",&node[i].v,&node[i].x);        max_size = max(max_size,node[i].x) + 1;    }    sort(node,node + n);    LL ans = 0;    for(int i = 0; i < n; i++){        int x = node[i].x,v = node[i].v;        ans += ((LL)(x * query(v2,1,max_size,1,x,1)) - (LL)(query(v1,1,max_size,1,x,1))) * v;        ans += ((LL)(query(v1,1,max_size,x,max_size,1)) - (LL)(query(v2,1,max_size,x,max_size,1) * x)) * v;        //printf("%d %I64d\n",i,ans);        update(v2,1,max_size,x,1,1);        update(v1,1,max_size,x,1,x);    }    printf("%I64d\n",ans);    return 0;}

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