算法課複習 -- 分治

HDU #5178 : pairs

傳送門：http://acm.hdu.edu.cn/showproblem.php?pid=5178

題意：n個數，問有多少對數差值小於k。

思路：排個序，開個隊列。每次一個新的數先把比它小的都從隊列中去了，然後答案加上隊列大小，最後把自己塞進隊列。按順序走一遍行了。

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const double _e = 10e-6;
const int maxn = 1e5 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

int x, y, z;
char c;

int t, n, k;
int a[maxn];
ll ans;

void solve()
{
	ans = 0;
	queue<int> q;
	for (int i = 1; i <= n; i++) {
		while (!q.empty() && a[i] - q.front() > k)
			q.pop();
		ans += 1ll * q.size();
		q.push(a[i]);
	}
}

int main()
{
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &k);
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		sort(a + 1, a + n + 1);
		solve();
		printf("%lld\n", ans);
	}
	return 0;
}

 

OpenJ_Bailian #2299 : Ultra-QuickSort

傳送門：http://bailian.openjudge.cn/practice/2299?lang=en_US

題意：給出一串數，每個數只能和相鄰的數交換位置，問最少交換多少次能使這一串數從小到大排序。

思路：求逆序對個數，樹狀數組直接做就行了。或者歸併排序來做也可以。

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const double _e = 10e-6;
const int maxn = 5e5 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

int x, y, z;
char c;

int lowbit(int x)
{
	return x&(-x);
}

int n, cnt;
int a[maxn], b[maxn];
map<int, int> mp;

void init()
{
	mp.clear();
}

void add(int x)
{
	for (int i = x; i <= cnt; i += lowbit(i))
		a[i]++;
}

int query(int x)
{
	int res = 0;
	for (int i = x; i > 0; i -= lowbit(i))
		res += a[i];
	return res;
}

int main()
{
	while (~scanf("%d", &n)) {
		if (n == 0)break;
		init();
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			b[i] = a[i];
		}
		sort(a + 1, a + n + 1);
		cnt = 0;
		for (int i = 1; i <= n; i++) {
			if (mp[a[i]] == 0)
				mp[a[i]] = ++cnt;
		}
		for (int i = 1; i <= n; i++)
			b[i] = mp[b[i]];

		cl0(a);
		ll ans = 0;
		for (int i = n; i > 0; i--) {
			ans += 1ll * query(b[i] - 1);
			add(b[i]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}

 

HDU #2083 : 簡易版之最短距離

傳送門：http://acm.hdu.edu.cn/showproblem.php?pid=2083

題意：給出n個點，選擇一個點使得所有點到它的距離之和最小，問最小距離和是多少。

思路：直接找最中間位置的點就可以了。

比如對於a、b兩個點（a<=b），選擇點c，若a<=c<=b則無論c在哪都有|ac|+|cb|=|b-a|；否則|ac|+|cb|>=|b-a|

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const double _e = 10e-6;
const int maxn = 1e5 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

int x, y, z;
char c;

int t, n;
int a[maxn];

int main()
{
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		sort(a + 1, a + n + 1);
		ll ans = 0;
		int l = 1, r = n;
		while (l < r) {
			ans += 1ll * (a[r] - a[l]);
			l++, r--;
		}
		printf("%lld\n", ans);
	}
	return 0;
}

 

OpenJ_Bailian #1064 : Cable master

傳送門：http://bailian.openjudge.cn/practice/1064?lang=en_US

題意：給n塊木頭，以及最終需要的塊數k。求出問把木頭切成k塊後，其中最短的那塊的最大值。

思路：二分答案。精度問題的話可以直接*100轉化成整數來求，或者二分個幾百次。

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9 + 10;
const double _e = 10e-6;
const int maxn = 1e4 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

int x, y, z;
char c;

int n, k;
double temp;
int a[maxn];

int main()
{
	while (~scanf("%d%d", &n, &k)) {
		for (int i = 1; i <= n; i++) {
			scanf("%lf", &temp);
			a[i] = temp * 100;
		}
		int l = 0, r = INF;
		while (l + 1 < r) {
			int mid = (l + r) / 2;
			int ans = 0;
			for (int i = 1; i <= n; i++)
				ans += a[i] / mid;
			if (ans >= k)
				l = mid;
			else
				r = mid;
		}
		int a1 = l / 100, a2 = l % 100;
		printf("%d.", a1);
		if (a2 < 10)printf("0");
		printf("%d\n", a2);
	}
	return 0;
}

 

OpenJ_Bailian #2456 : Aggressive cows

傳送門：http://bailian.openjudge.cn/practice/2456?lang=en_US

題意：有n個牛棚，c頭牛。牛比較兇猛所以它們太近了要打架。給出n個牛棚的位置，每個牛棚放一隻牛，問兩隻牛之間的最短距離的最大值是多少。

思路：二分答案。默認1號位置放牛，之後隔着dis位置的牛棚放牛，看放不放的下。

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const double _e = 10e-6;
const int maxn = 1e5 + 10;
const int matSize = 9;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };

int x, y, z;
char c;

int n, k;
int a[maxn];

bool solve(int dis)
{
	int pos = 1, cnt = k - 1;
	while (cnt > 0) {
		pos = lower_bound(a + 1, a + n + 1, a[pos] + dis) - a;
		if (pos > n)return false;
		cnt--;
	}
	return true;
}

int main()
{
	while (~scanf("%d%d", &n, &k)) {
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		sort(a + 1, a + n + 1);
		int l = 0, r = INF;
		while (l + 1 < r) {
			int mid = (l + r) / 2;
			if (solve(mid))
				l = mid;
			else
				r = mid;
		}
		printf("%d\n", l);
	}
	return 0;
}