hdu 2141 二分查找

直接枚举Ai, Bj, Ck的复杂度为L×M×N，过不了。先把所有Ai, Bj的和存起来并排序，然后枚举Ck，二分查找Ai, Bj的和，复杂度为L×M×log(L×M)。

/*
 * hdu2141/linux.cpp
 * Created on: 2011-9-1
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXN = 510;

int L, N, M, LM;
int A[MAXN], B[MAXN], C[MAXN], sum[MAXN * MAXN];

bool find(int num) {
    int mid, low, high;
    low = 0;
    high = LM - 1;
    while (low <= high) {
        mid = (low + high) / 2;
        if (sum[mid] == num) {
            return true;
        } else if (sum[mid] > num) {
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return low < LM && sum[low] == num;
}

bool search(int x) {
    for (int i = 0; i < N; i++) {
        if (find(x - C[i])) {
            return true;
        }
    }
    return false;
}

void work();
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    work();
    return 0;
}

void work() {
    int T = 0, S, x;
    while (scanf("%d%d%d", &L, &M, &N) == 3) {
        for (int i = 0; i < L; i++) {
            scanf("%d", &A[i]);
        }
        for (int i = 0; i < M; i++) {
            scanf("%d", &B[i]);
        }
        for (int i = 0; i < N; i++) {
            scanf("%d", &C[i]);
        }
        scanf("%d", &S);
        sort(A, A + L);
        sort(B, B + M);
        sort(C, C + N);
        printf("Case %d:\n", ++T);
        LM = 0;
        for (int i = 0; i < L; i++) {
            for (int j = 0; j < M; j++) {
                sum[LM++] = A[i] + B[j];
            }
        }
        sort(sum, sum + LM);
        while (S--) {
            scanf("%d", &x);
            if (search(x)) {
                puts("YES");
            } else {
                puts("NO");
            }
        }
    }
}