大连2011ACM网络赛【5道水题总结】……很黄很暴力

[size=medium]KIDx 的解题报告[/size]

[img]http://dl.iteye.com/upload/attachment/549202/25e5a98b-f15a-3820-93b6-bb9278333409.png[/img]

[url]http://acm.hdu.edu.cn/listproblem.php?vol=31[/url]
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4001：直接一个最长递增子序列模板，注意数据范围就可以了
[img]http://dl.iteye.com/upload/attachment/591362/5864fbcb-5570-3b6d-bf92-2069bd4f2cba.png[/img]
[/size][/b]

#include 
#include 
using namespace std;
#define L __int64
#define M 1005

struct block{
    L a, b, c, d;
}x[M];
L dp[M];

bool cmp (block x, block y)
{
    if (x.a == y.a)
    {
        if (x.b == y.b)
            return x.d > y.d;
        return x.b < y.b;
    }
    return x.a < y.a;
}
int main()
{
    int n, i, j;
    while (scanf ("%d", &n), n)
    {
        for (i = 0; i < n; i++)
        {
            scanf ("%I64d%I64d%I64d%I64d", &x[i].a, &x[i].b, &x[i].c, &x[i].d);
            if (x[i].a < x[i].b)
                x[i].a ^= x[i].b, x[i].b ^= x[i].a, x[i].a ^= x[i].b;
        }
        sort (x, x+n, cmp);
        for (i = 0; i < n; i++)
            dp[i] = x[i].c;
        for (i = 0; i < n; i++)
        {
            for (j = i + 1; j < n; j++)
            {
                if (x[j].d == 0 && x[j].a >= x[i].a && x[j].b >= x[i].b ||
                    x[j].d == 1 && x[j].a >= x[i].a && x[j].b >= x[i].b && 
x[j].a*x[j].b > x[i].a*x[i].b ||
                    x[j].d == 2 && x[j].a > x[i].a && x[j].b > x[i].b)
                    if (dp[j] < dp[i]+x[j].c)
                        dp[j] = dp[i]+x[j].c;
            }
        }
        printf ("%I64d\n", *max_element (dp, dp+n));
    }
    return 0;
}

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4002：猥琐打表找规律+大数乘法，Java暴力水过
[img]http://dl.iteye.com/upload/attachment/591364/ac5b9275-7a77-3c93-b1e7-09dfdb3877a4.png[/img]
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import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
    static public void main(String[] args) throws IOException
    {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        BigInteger prime[] = new BigInteger[60], tp, n;
        int hash[] = new int[300], k, i, j, num, t;
        for (i = 0; i < 300; i++)
            hash[i] = 1;
        k = 0;
        for (i = 2; i < 242; i++)
        {
            if (hash[i] == 1)
            {
                prime[k] = BigInteger.valueOf(i);
                k++;
                for (j = i * 2; j < 242; j += i)
                    hash[j] = 0;
            }
        }
        for (i = 1; i < k; i++)
            prime[i] = prime[i].multiply(prime[i-1]);
        t = cin.nextInt();
        while (true)
        {
            if (t == 0)
                break;
            t--;
            n = cin.nextBigInteger();
            for (i = 0; i < k; i++)
                if (n.compareTo(prime[i]) == -1)
                    break;
            System.out.println(prime[i-1]);
        }
    }
}

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4004：二分枚举答案+暴力检验
[img]http://dl.iteye.com/upload/attachment/591366/57e9d60f-f075-34e3-b436-d6e6538d00d8.png[/img]
[/size][/b]

#include 
#include 
using namespace std;
#define M 500005

int x[M], v[M];

int main()
{
	int L, n, m, i, k, maxs, l, r, mid, num, tp;
	while (~scanf ("%d%d%d", &L, &n, &m))
	{
		x[0] = 0;
		for (i = 1; i <= n; i++)
			scanf ("%d", x+i);
		sort (x, x+n+1);
		x[n+1] = L;
		k = maxs = 0;
		for (i = 1; i < n + 2; i++)
		{
			v[k++] = x[i] - x[i-1];
			if (v[k-1] > maxs)
				maxs = v[k-1];
		}
		l = maxs, r = L;
		while (l < r)
		{
			tp = num = 0;
			mid = (l + r) / 2;
			for (i = 0; i < k; i++)
			{
				if (tp + v[i] > mid)
					num++, tp = v[i];
				else tp += v[i];
			}
			num++;
			if (num <= m)
				r = mid;
			else l = mid + 1;
		}
		printf ("%d\n", r);
	}
    return 0;
}

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4006：优先队列暴力水过
[img]http://dl.iteye.com/upload/attachment/591360/7a35027c-acc9-3d5a-a316-ee8cfaf07778.png[/img]
[/size][/b]

#include 
#include 
using namespace std;

struct Int{
    int v;
    friend bool operator < (Int a, Int b)
    {
        return a.v > b.v;
    }
};

int main()
{
    Int tp;
    int n, x, k;
    char ch[3];
    while (~scanf ("%d%d", &n, &k))
    {
        priority_queue q;
        while (n--)
        {
            scanf ("%s", ch);
            if (ch[0] == 'Q')
            {
                while (q.size() > k)
                    q.pop();
                printf ("%d\n", q.top().v);
            }
            else
            {
                scanf ("%d", &x);
                tp.v = x;
                q.push (tp);
            }
        }
    }
    return 0;
}

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4007：最暴力的解法，相信没有人能够破我记录----史上最慢[img]http://dl.iteye.com/upload/attachment/549230/2551a8b6-cdbd-3bc2-b680-5a2597cc3030.gif[/img]
[img]http://dl.iteye.com/upload/attachment/591358/f1dde3c2-5716-3124-b737-cc5f68443616.png[/img]
先优先sort-x坐标，再枚举2条垂直于x轴的扫描线，再从p数组中筛选出在这2条扫描线中的x个点入tp数组，然后优先sort-y坐标，再枚举2条垂直于y轴的扫描线，再从tp数组中筛选出在这2条扫描线中的tmp个点，就是4条扫描线所围成的正方形里的点的个数
[/size][/b]

#include 
#include 
using namespace std;
#define M 1005

struct point{
    int x, y;
}p[M], tp[M];

bool cmp (point a, point b)
{
    if (a.x == b.x)
        return a.y < b.y;
    return a.x < b.x;
}
bool cmp2 (point a, point b)
{
    if (a.y == b.y)
        return a.x < b.x;
    return a.y < b.y;
}
int main()
{
    int n, R, i, j, k, lx, rx, ly, ry, x, maxs, tmp;
    while (~scanf ("%d%d", &n, &R))
    {
        for (i = 0; i < n; i++)
            scanf ("%d%d", &p[i].x, &p[i].y);
        sort (p, p+n, cmp);
        maxs = 0;
        for (i = 0; i < n; i++)
        {
            rx = p[i].x;
            lx = rx - R;
            x = 0;
            for (j = 0; j < n; j++)
            {
                if (p[j].x > rx)
                    break;
                if (p[j].x >= lx && p[j].x <= rx)
                    tp[x++] = p[j];
            }
            sort (tp, tp+x, cmp2);
            for (j = 0; j < x; j++)
            {
                ry = tp[j].y;
                ly = ry - R;
                tmp = 0;
                for (k = 0; k < x; k++)
                {
                    if (tp[k].y > ry)
                        break;
                    if (tp[k].y >= ly && tp[k].y <= ry)
                        tmp++;
                }
                if (tmp > maxs)
                    maxs = tmp;
            }
        }
        printf ("%d\n", maxs);
    }
    return 0;
}