CF 431C k-Tree

2/8/2018
Problem: http://codeforces.com/problemset/problem/431/C
Solution: DP. We can generalie the problem into, divided number n as the sum of a couple numbers less than k. All so, one of them must be greater or equal to d.
So, we construct a 2-D dp with 0 or 1 denoting whether there already is one number greater or equal to d as the 2nd dimension.

#include
using namespace std;

int main() {
    int n, k, d;
    cin >> n >> k >> d;
    long long dp[105][2] = {0};
    long long mod = 1e9 + 7;
    dp[0][0] = 1;
    dp[0][1] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            if (i < j) break;
            if (j < d) {
                dp[i][0] = (dp[i][0] + dp[i - j][0]) % mod;
                dp[i][1] = (dp[i][1] + dp[i - j][1]) % mod;
            } else {
                dp[i][1] = (dp[i][1] + dp[i - j][0] + dp[i - j][1]) % mod;
            }
        }
    }
    cout << dp[n][1];
    return 0;
}

POJ 3984/ kuangbin 1K/ maze problem

Problem: http://poj.org/problem?id=3984
Solution: BFS.

#include
#include
#include
#include
using namespace std;

int a[6][6] = {0};
int vis[6][6] = {0};
int dx[5] = {0, -1, 0, 1, 0};
int dy[5] = {0, 0, -1, 0, 1};
int prevx[6][6] = {0};
int prevy[6][6] = {0};

struct T {
	int x;
	int y;
	void init(int a, int b) {
		x = a;
		y = b;
	}
};

queue q;
stack ans;

void print(int x, int y) {
	if (x != 0 || y != 0) {
		T temp;
		temp.init(x, y);
		ans.push(temp);
		print(prevx[x][y], prevy[x][y]);
	}
}

void bfs(T temp) {
	int m = temp.x;
	int n = temp.y;
	if (m == 5 && n == 5) {
		print(m, n);
	} else {
		q.pop();
		for (int i = 1; i = 1 && m + dx[i] = 1 && n + dy[i] <= 5 && 
						a[m + dx[i]][n + dy[i]] == 0 && vis[m + dx[i]][n + dy[i]] == 0) {
				vis[m + dx[i]][n + dy[i]] = 1;
				T temp;
				temp.init(m + dx[i], n + dy[i]);
				prevx[m + dx[i]][n + dy[i]]	= m;
				prevy[m + dx[i]][n + dy[i]] = n;
				q.push(temp);
			}
		}
		bfs(q.front());
	}
}

int main () {
	for (int i = 1; i  a[i][j];
		}
	}
	T temp;
	temp.init(1, 1);
	prevx[temp.x][temp.y] = 0;
	prevy[temp.x][temp.y] = 0;
	vis[temp.x][temp.y] = 1;
	q.push(temp);
	bfs(temp);
	while(!ans.empty()) {
		cout << '(' << ans.top().x - 1 << ", " << ans.top().y - 1 << ')';
		ans.pop();
		if (!ans.empty()) {
			cout << endl;
		}
	}
	return 0;
}

POJ 3126 Prime Path/ kuangbin 1F

12/23/2017
Problem: http://poj.org/problem?id=3126
Solution: BFS.

#include<iostream>
#include<queue>
#include<string.h> 
using namespace std;
int mp[10000];

struct node {
	int num, cnt;
	node(int numm, int cntt) {
		num = numm;
		cnt = cntt;
	}
};

queue<node> q;

bool prime(int n) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			return false;
		}
	}
	return true;
} 

void digit(node n) {
	int div = 0, rem = 0;
	int tmp = 1;	
	for (int i = 1; i <= 3; i++) {
		tmp *= 10;
		div = n.num / tmp;
		rem = n.num % (tmp / 10);
		for (int j = 0; j <= 9; j++) {
			int newNum = div * tmp + j * (tmp / 10) + rem;
			// cout << newNum << endl; 
			if (prime(newNum) && !mp[newNum]) {
				// cout << newNum << endl;
				q.push(node(newNum, n.cnt + 1)); 
				mp[newNum] = 1;
			} 
		}
	}
	rem = n.num % 1000;
	for (int j = 1; j <= 9; j++) {
		int newNum = j * 1000 + rem;
		if (prime(newNum) && !mp[newNum]) {
			// cout << newNum << endl; 
			q.push(node(newNum, n.cnt + 1));
			mp[newNum] = 1;
		}
	}
}

int main() {
	int n;
	cin >> n;
	while (n--) {
		int start = 0, end = 0;
		cin >> start >> end;
		memset(mp, 0, sizeof(mp));
		while (!q.empty()) {
			q.pop();
		}
		node head = node(start, 0);
		mp[start] = 1;
		q.push(head);
		while (!q.empty() && q.front().num != end) {
			digit(q.front());
			q.pop();
		}
		if (q.empty()) {
			cout << "Impossible" << endl; 
		} else {
			cout << q.front().cnt << endl;			
		}
	}
	return 0;
}

hdu 1272/ kuangbin 5-M(graph)

10/17/2017
Problem: http://acm.split.hdu.edu.cn/showproblem.php?pid=1272
Solution: Use the identity that for a tree, the number of nodes should be one
more than that of edges. Pay special attention to that there could be
zero node, but the answer should still be true.
Note: Another possible solution, using dsu, is posted here: https://prionanthaspace.wordpress.com/2017/10/18/hdu-1272-kuangbin-5-m/

#include<iostream>
#include<string.h>
using namespace std;

int flag[100005] = {0};

int main() {
	int a, b;
	int cntn = 0, cnte = 0;
	while(cin >> a >> b) {
		if (a == -1 && b == -1) {
			return 0;
		} else if (!a && !b) {
			bool ans = (cntn == cnte + 1) || cntn == 0; 
			cout << (ans? "Yes": "No") << endl;
			memset(flag, 0, sizeof(flag));
			cntn = 0;
			cnte = 0;
			continue;
		}
		cnte ++;
		if (!flag[a]) {
			flag[a] = 1;
			cntn ++;
		}
		if (!flag[b]) {
			flag[b] = 1;
			cntn ++;
		}
	}
	return 0;
} 

hdu 1272/ kuangbin 5-M(dsu)

10/17/2017
Problem: http://acm.split.hdu.edu.cn/showproblem.php?pid=1272
Solution: DSU. However, also need to check if all dots are connected. For that,
simply loop every dot that exists, and check if all their father are
the same.
Note: This problem would be much more easier if we consider the identity of a
tree, which is the number of dot is always one more than that of edges.
Link: https://prionanthaspace.wordpress.com/2017/10/18/hdu-1272-kuangbin-5-mgraph/

#include<iostream>
#include<string.h>
#include<cmath>
using namespace std;

int parent[100005] = {0};
int flag[100005] = {0};

int find(int a) {
	if (parent[a] == a) {
		return a;
	} else {
		return parent[a] = find(parent[a]);
	}
}

void merge(int a, int b) {
	int p1 = find(a);
	int p2 = find(b);
	if (p1 != p2) {
		parent[p1] = p2;
	}
}

int main() {
	int a, b;
	bool ans = true;
	int maxx = 0;
	int minn = 100005;
	for (int i = 1; i <= 100005; i++) {
		parent[i] = i;
	}	
	memset(flag, 0, sizeof(flag));
	while (cin >> a >> b) {
		if (a == -1 && b == -1) {
			return 0;
		} else if (!a && !b) {
			if (ans) {
				int tmp = find(minn);
				for (int i = minn + 1; i <= maxx; i++) {
					if (flag[i] && find(i) != tmp) {
						// cout << tmp << ' ' << find(i) << endl;
						ans = false;
						break;
					}
				}
			}		
			cout << (ans? "Yes" :"No") << endl;
			maxx = 0;
			minn = 100005;			
			ans = true;
			for (int i = 1; i <= 100005; i++) {
				parent[i] = i;
			}
			memset(flag, 0, sizeof(flag));
			continue;
		}
		minn = min(minn, min(a, b));
		maxx = max(maxx, max(a, b));
		flag[a] = 1;
		flag[b] = 1;
		if (find(a) != find(b)) {
			merge(a, b);
		} else {
			ans = false;
		}
	}
	return 0;
}

HDU 1213/ kuangbin 5-C

10/16/2017
Problem: http://acm.split.hdu.edu.cn/showproblem.php?pid=1213

#include<iostream>
#include<string.h>
using namespace std;

int parent[1005] = {0};
int cnt = 0;

int find(int a) {
	if (parent[a] == a) {
		return a;
	} else {
		return parent[a] = find(parent[a]);
	}
}

void merge(int a, int b) {
	int p1 = find(a);
	int p2 = find(b);
	if (p1 != p2) {
		parent[p1] = p2;
		cnt --;
	}
}

int main() {
	int t;
	cin >> t;
	while (t--) {
		cnt =0;
		int n, m;
		cin >> n >> m;
		cnt = n;
		for (int i = 1; i <= n; i++) {
			parent[i] = i;
		}
		for (int i = 1; i <= m; i++) {
			int a, b;
			cin >> a >> b;
			if (find(a) != find(b)) {
				merge(a, b);
			}
		}
		cout << cnt << endl;
	}
	return 0;
} 

POJ 1426/ kuangbin 1E

10/16/2017
Problem: http://poj.org/problem?id=1426
Solution: BFS. Manage a queue, and for each time push current number + 0 or 1 in
the end all the way untill curr is a multiple of given n. For some
mysterious reason limited to long long…

#include<iostream>
#include<queue>
using namespace std;

int main() {
	long long n;
	while (cin >> n) {
		if (n == 0) {
			break;
		}
		long long ans = 0;
		queue<long long> q;
		q.push(1);
		while (!q.empty()) {
			long long cur = q.front();
			q.pop();
			if (cur % n == 0) {
				cout << cur << endl;
				break;
			} else {
				q.push(cur * 10);
				q.push(cur * 10 + 1);
			}
		}
	}
	return 0;
} 

POJ 3278/ kuangbin 1C

10/13/2017
Problem: http://poj.org/problem?id=3278
Solution: For every element x in the queue, add in x-1. x+1, x*2.

#include<iostream>
#include<queue>
using namespace std;

bool v[100005];

bool inrange(int x) {
	return (x >= 0) && (x <= 100000); 
}

struct node{
	int t;
	int x;
	node(int tt, int xx) {
		t = tt;
		x = xx;
	}
};

int main() {
	int n, k;
	cin >> n >> k;
	queue<node> q;
	q.push(node(0, n));
	v[n] = true;
	while(!q.empty()) {
		node curr = q.front();
		q.pop();
		if (curr.x == k) {
			cout << curr.t;
			return 0;
		}
		if (inrange(curr.x - 1) && !v[curr.x - 1]) {
			q.push(node(curr.t + 1, curr.x - 1));
			v[curr.x - 1] = true;
		} 
		if (inrange(curr.x + 1) && !v[curr.x + 1]) {
			q.push(node(curr.t + 1, curr.x + 1));
			v[curr.x + 1] = true;
		}
		if (inrange(curr.x * 2) && !v[curr.x * 2]) {
			q.push(node(curr.t + 1, curr.x * 2));
			v[curr.x * 2] = true;
		}
	}
	return 0;
} 

POJ 2251 Dungeon Master/ kuangbin 1B

10/13/2017
Problem: http://poj.org/problem?id=2251
Solution: A plain bfs searching. Only changing 2D to 3D, but the usage is the strategy is the same.

#include<iostream>
#include<string.h>
#include<cstring>
#include<queue>
using namespace std;

int x, y, z;
int xs, ys, zs, xe, ye, ze;
int step = 0;
char map[35][35][35];
bool v[35][35][35];

int dx[10] = {1, 0, 0, -1, 0, 0}; 
int dy[10] = {0, 1, 0, 0, -1, 0}; 
int dz[10] = {0, 0, 1, 0, 0, -1}; 

struct node{
	int time;
	int x, y, z;
	node(int tt, int xx, int yy, int zz) {
		time = tt;
		x = xx;
		y = yy;
		z = zz;
	}
};

bool inrange(int xx, int yy, int zz) {
	return (xx >= 0 && yy >= 0 && zz >= 0 && xx < x && yy < y && zz < z);
}

bool bfs() {
	queue<node> q;
	q.push(node(0, xs, ys, zs));
	v[xs][ys][zs] = true;
	while (!q.empty()) {
		node curr = q.front();
		q.pop();
		if (curr.x == xe && curr.y == ye && curr.z == ze) {
			step = curr.time;
			return true;
		} else {
			for (int i = 0; i < 6; i++) {
				int xx = curr.x + dx[i];
				int yy = curr.y + dy[i];
				int zz = curr.z + dz[i];
				if (inrange(xx, yy, zz) && (map[xx][yy][zz] == '.' || map[xx][yy][zz] == 'E') && !v[xx][yy][zz]) {
					q.push(node(curr.time + 1, xx, yy, zz));
					v[xx][yy][zz] = true;
				}
			}
		}
	}
	return false;
}

int main() {
	while (cin >> x >> y >> z) {
		if (x == 0 && y == 0 && z == 0) {
			return 0;
		}
		memset(map, 0, sizeof(map));
		memset(v, 0, sizeof(v));
		for (int i = 0; i < x; i++) {
			for (int j = 0; j < y; j++) {
				cin >> map[i][j];
				for (int k = 0; k < z; k++) {
					if (map[i][j][k] == 'S') {
						xs = i;
						ys = j;
						zs = k;
					} else if (map[i][j][k] == 'E') {
						xe = i;
						ye = j;
						ze = k;
					}
				}
			}
		}
		if (bfs()) {
			cout << "Escaped in " << step << " minute(s).";
		} else {
			cout << "Trapped!";
		}
		cout << endl;
	}
	return 0;
}

POJ 2349

8/8/2017
Problem: http://poj.org/problem?id=2349
Solution: The answer is just the length of the k-th largest edge in the MST.

#include<cstdio>
#include<iostream> 
#include<vector>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

struct point{
	int x, y;
	point(int xx, int yy):x(xx), y(yy){}
	point(){}
};

struct edge{
	int i, j;
	double w;
	edge(int ii, int jj, double ww):i(ii), j(jj), w(ww){}
	edge(){}
};

point p[5005];
edge e[251010];
int parent[5005];

double dist(int x1, int y1, int x2, int y2) {
	return sqrt((double)(x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}

bool cmp(edge a, edge b) {
	return a.w < b.w;
}

int find(int a) {
	if (parent[a] == a) {
		return a;
	} else {
		parent[a] = find(parent[a]);
		return parent[a];
	}
}

void merge(int a, int b) {
	int p1 = find(a);
	int p2 = find(b);
	if (p1 != p2) {
		parent[p2] = p1;
	}
}

int main() {
	int t;
	cin >> t;
	while (t--) {
		int s, n;
		cin >> s >> n;
		for (int i = 1; i <= n; i++) {
			parent[i] = i;
		} 
		for (int i = 1; i <= n; i++) {
			cin >> p[i].x >> p[i].y;
		} 
		int cnt = 0;
		for (int i = 1; i < n; i++) {
			for (int j = i + 1; j <= n; j++) {
				double tmp = dist(p[i].x, p[i].y, p[j].x, p[j].y);
				e[++cnt] = edge(i, j, tmp);
				e[++cnt] = edge(j, i, tmp);
			}
		}
		int node = 0;
		sort(e + 1, e + cnt + 1, cmp);
		for (int i = 1; i <= cnt; i++) {
			if (find(e[i].i) != find(e[i].j)) {
				merge(e[i].i, e[i].j);
				node++;
				if (node == n - s) {
					printf("%0.2f\n", e[i].w);
					break;
				}
			}
		}
	}
	return 0;
}