UVA 1152 4 Values Whose Sum is Zero 和为0的4个值

摘要：中途相遇。对比map，快排+二分查找，Hash效率。

n是4000的级别，直接O（n^4）肯定超，所以中途相遇法，O（n^2）的时间枚举其中两个的和，O（n^2）的时间枚举其他两个的和的相反数，然后O（logN）的时间查询是否存在。

首先试了下map，果断TLE

//TLE
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;

const int maxn = 4001;
int data[4][maxn];

map<int,int> cnt;

int main()
{

    int T ; scanf("%d",&T);
    int *A = data[0], *B = data[1], *C = data[2],*D = data[3];
    map<int,int>::iterator it;
    while(T--){
        int n; scanf("%d",&n);
        for(int i = 0; i < n; i++){
            scanf("%d%d%d%d",A+i,B+i,C+i,D+i);
        }

        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            cnt[A[i]+B[j]]++;
        }

        int ans = 0;
        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            int tmp = -C[i]-D[j];
            it = cnt.find(tmp);
            if(it!=cnt.end()) ans += it->second;
        }
        printf("%d\n",ans);
        if(T) putchar(‘\n‘);
    }

    return 0;
}

map，TLE

然后改成了快排+二分查找，4920ms

// runtime 4920ms
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 4001;
int data[4][maxn];
int vec[maxn*maxn];

int _lower_bound(int *A,int L,int R,int v)
{
    int m;
    while(L<R){
        m = (L+R)>>1;
        if(A[m]>=v) R = m;
        else L = m+1;
    }
    return L;
}

int _upper_bound(int *A,int L,int R,int v)
{
    int m;
    while(L<R){
        m = (L+R)>>1;
        if(A[m]>v) R = m;
        else L = m+1;
    }
    return L;
}

int main()
{

    int T ; scanf("%d",&T);
    int *A = data[0], *B = data[1], *C = data[2],*D = data[3];
    while(T--){
        int n; scanf("%d",&n);
        for(int i = 0; i < n; i++){
            scanf("%d%d%d%d",A+i,B+i,C+i,D+i);
        }

        int sz = 0;
        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            vec[sz++] = A[i]+B[j];
        }
        sort(vec,vec+sz);

        int ans = 0;
        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            int tmp = -(C[i]+D[j]);
            ans +=  _upper_bound(vec,0,sz,tmp) - _lower_bound(vec,0,sz,tmp);
        }
        printf("%d\n",ans);
        if(T) putchar(‘\n‘);
    }

    return 0;
}

快拍+二分，4920ms

更好一点的做法快排+计数，2832ms

#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
typedef pair<int,int> pii;
#define fi first
#define se second
const int maxn = 4001;
int data[4][maxn];

pii cnt1[maxn*maxn];
pii cnt2[maxn*maxn];
int vec[maxn*maxn];

int main()
{
    int T ; scanf("%d",&T);
    int *A = data[0], *B = data[1], *C = data[2],*D = data[3];
    while(T--){
        int n; scanf("%d",&n);
        for(int i = 0; i < n; i++){
            scanf("%d%d%d%d",A+i,B+i,C+i,D+i);
        }
        int sz = 0;
        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            vec[sz++] = A[i]+B[j];
        }
        sort(vec,vec+sz);
        int len1 = 0;
        cnt1[len1] = pii(vec[len1],1);
        for(int i = 1; i < sz; i++){
            if(vec[i] == cnt1[len1].fi) cnt1[len1].se++;
            else { cnt1[++len1].fi = vec[i]; cnt1[len1].se = 1; }
        }
        sz = 0;
        for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            vec[sz++] = -C[i]-D[j];
        }
        sort(vec,vec+sz);
        int len2 = 0;
        cnt2[len2] = pii(vec[len2],1);
        for(int i = 1; i < sz; i++){
            if(vec[i] == cnt2[len2].fi) cnt2[len2].se++;
            else { cnt2[++len2].fi = vec[i]; cnt2[len2].se = 1; }
        }

        int p = 0,q = 0,ans = 0;
        while(p<=len1&&q<=len2){
            if(cnt1[p].fi == cnt2[q].fi){
                ans += cnt1[p++].se*cnt2[q++].se;
            }else if(cnt1[p].fi>cnt2[q].fi) q++;
                  else p++;
        }
        printf("%d\n",ans);
        if(T) putchar(‘\n‘);
    }

    return 0;
}

还有Hash表，写挂了，待补。。。