hdu2795

maksyuki 发表于 oj 分类，标签: 数据结构-线段树

18 9月 2015

Billboard

Problem Description

At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.
On September 1, the billboard was empty. One by one, the announcements started being put on the billboard. Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi. When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one. If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university). Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

Input

There are multiple cases (no more than 40 cases). The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements. Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.

Output

For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.

Sample Input

3 5 5
2
4
3
3
3

Sample Output

-1

Author

hhanger@zju

Source

HDOJ 2009 Summer Exercise（5）

题目类型：线段树+离散化

算法分析：n的规模只有2e5，则易知线段树中的点的个数最多也就是n的。然后将wi插入到树中，然后更新最值即可，线段树的插入操作本身就能够保证当前wi是插入到符合的位置上的

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 2e5 + 66;

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r


LL maxval[maxn<<2];

void PushUp (int rt)
{
    maxval[rt] = max (maxval[rt<<1], maxval[rt<<1|1]);
}

LL Query (int rt, int l, int r, int v)
{
    if (l == r)
    {
        maxval[rt] -= v;
        return l;
    }

    int m = (l + r) >> 1;
    LL tt;
    if (v <= maxval[rt<<1])
        tt = Query (lson, v);

    else
        tt = Query (rson, v);

    PushUp (rt);
    return tt;
}


int main()
{
//    CPPFF;
    int h, w, q;
    while (scanf ("%d%d%d", &h, &w, &q) != EOF)
    {
        fill (maxval, maxval + (maxn << 2), w);
        if (h > q) h = q;

        for (int i = 1; i <= q; i++)
        {
            int wi;
            scanf ("%d", &wi);
            if (wi > maxval[1]) puts ("-1");

            else
                printf ("%lld\n", Query (1, 1, h, wi));
        }
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 2e5 + 66;

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

LL maxval[maxn<<2];

void PushUp (int rt)

{

maxval[rt] = max (maxval[rt<<1], maxval[rt<<1|1]);

}

LL Query (int rt, int l, int r, int v)

{

if (l == r)

{

maxval[rt] -= v;

return l;

}

int m = (l + r) >> 1;

LL tt;

if (v <= maxval[rt<<1])

tt = Query (lson, v);

else

tt = Query (rson, v);

PushUp (rt);

return tt;

}

int main()

{

// CPPFF;

int h, w, q;

while (scanf ("%d%d%d", &h, &w, &q) != EOF)

{

fill (maxval, maxval + (maxn << 2), w);

if (h > q) h = q;

for (int i = 1; i <= q; i++)

{

int wi;

scanf ("%d", &wi);

if (wi > maxval[1]) puts ("-1");

else

printf ("%lld\n", Query (1, 1, h, wi));

}

return 0;

}

hdu1394

maksyuki 发表于 oj 分类，标签: 数据结构-线段树

18 9月 2015

Minimum Inversion Number

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

Output

For each case, output the minimum inversion number on a single line.

Sample Input

1 3 6 9 0 8 5 7 4 2

Sample Output

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

题目类型：线段树

算法分析：先求出初始序列的逆序对数，然后对于后面的序列可以通过递推得到，这是因为每次都是将第一个数取出来放到最后，则可以知道取出这个数会使总的逆序对数损失aa[i]个，而放到最后会使总的逆序对数增加n - aa[i] - 1个

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e4 + 66;

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

LL sum[maxn<<2];

void PushUp (int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void UpDate (int rt, int l, int r, int v)
{
    if (l == r)
    {
        sum[rt]++;
        return ;
    }

    int m = (l + r) >> 1;
    if (v <= m)
        UpDate (lson, v);

    else
        UpDate (rson, v);

    PushUp (rt);
}

LL Query (int rt, int l, int r, int L, int R)
{
    if (L <= l && r <= R)
        return sum[rt];

    int m = (l + r) >> 1;
    LL res = 0;
    if (L <= m)
        res += Query (lson, L, R);

    if (m < R)
        res += Query (rson, L, R);

    return res;
}

int aa[maxn];

int main()
{
//    CPPFF;
    int n;
    while (cin >> n)
    {
        LL res = 0, minval = INF;
        memset (sum, 0, sizeof (sum));
        for (int i = 1; i <= n; i++)
        {
            cin >> aa[i];
            res += Query (1, 0, n - 1, aa[i] + 1, n - 1);
            UpDate (1, 0, n - 1, aa[i]);
        }
        minval = min (minval, res);

        for (int i = 1; i <= n; i++)
        {
            res += n - aa[i] - aa[i] - 1;
            minval = min (minval, res);
        }
        cout << minval << endl;
    }
    return 0;
}

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e4 + 66;

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

LL sum[maxn<<2];

void PushUp (int rt)

{

sum[rt] = sum[rt<<1] + sum[rt<<1|1];

}

void UpDate (int rt, int l, int r, int v)

{

if (l == r)

{

sum[rt]++;

return ;

}

int m = (l + r) >> 1;

if (v <= m)

UpDate (lson, v);

else

UpDate (rson, v);

PushUp (rt);

}

LL Query (int rt, int l, int r, int L, int R)

{

if (L <= l && r <= R)

return sum[rt];

int m = (l + r) >> 1;

LL res = 0;

if (L <= m)

res += Query (lson, L, R);

if (m < R)

res += Query (rson, L, R);

return res;

}

int aa[maxn];

int main()

{

// CPPFF;

int n;

while (cin >> n)

{

LL res = 0, minval = INF;

memset (sum, 0, sizeof (sum));

for (int i = 1; i <= n; i++)

{

cin >> aa[i];

res += Query (1, 0, n - 1, aa[i] + 1, n - 1);

UpDate (1, 0, n - 1, aa[i]);

}

minval = min (minval, res);

for (int i = 1; i <= n; i++)

{

res += n - aa[i] - aa[i] - 1;

minval = min (minval, res);

}

cout << minval << endl;

}

return 0;

}

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超客的水墨烟雨

hdu2795

Billboard

hdu1394

Minimum Inversion Number

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