算法課複習 -- 圖

hihoCoder #1322 : 樹結構判定

傳送門：https://hihocoder.com/problemset/problem/1322

題意：給出一個n個頂點m條邊的無向圖，問是不是樹。

思路：首先作爲一棵樹必然滿足n=m+1，若滿足則任取一點dfs判斷有沒有圈，最後看是不是所有點都遍歷到了。

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 int N = 410;
const double _e = 10e-6;
const int maxn = 2e5;
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;
int t, n, m;
vector<int> G[510];
bool judge;
bool used[510];

void dfs(int u, int fa)
{
	used[u] = true;
	for (int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];
		if (fa == v)continue;
		if (used[v]) {
			judge = true;
			return;
		}
		dfs(v, u);
		if (judge)return;
	}
}

int main()
{
	while (~scanf("%d", &t)) {
		while (t--) {
			judge = false; cl0(used); 
			scanf("%d%d", &n, &m);
			for (int i = 1; i <= n; i++)
				G[i].clear();
			for (int i = 1; i <= m; i++) {
				scanf("%d%d", &x, &y);
				G[x].push_back(y); G[y].push_back(x);
			}
			if (n != m + 1) {
				puts("NO");
				continue;
			}
			dfs(1, -1);
			for (int i = 1; i <= n; i++) {
				if (!used[i]) {
					judge = true;
					break;
				}
			}
			if (judge) puts("NO");	
			else puts("YES");
		}
	}
	return 0;
}

 

CodeChef #SCC5 : Celebrity In Trouble

傳送門：https://www.codechef.com/problems/SCC5

題意：有n個人，如果一個人不認識任何其他的人，並且其他人都認識他，那麼他被稱作Celebrity，問是否存在這樣的人。

思路：記入度和出度，入度爲n-1並且出度爲0的人就是。

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 int N = 410;
const double _e = 10e-6;
const int maxn = 2e5;
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;
int t, n, m;
int in[1010], out[1010];

int main()
{
	while (~scanf("%d", &t)) {
		while (t--) {
			bool judge = false;
			cl0(in); cl0(out);
			scanf("%d%d", &n, &m);
			for (int i = 1; i <= m; i++) {
				scanf("%d%d", &x, &y);
				in[y]++; out[x]++;
			}
			for (int i = 1; i <= n; i++) {
				if (in[i] == n - 1 && out[i] == 0) {
					printf("alive %d\n", i);
					judge = true;
					break;
				}
			}
			if (!judge)
				puts("dead");
		}
	}
	return 0;
}

 

POJ #1386 : Play on Words

傳送門：http://poj.org/problem?id=1386

題意：給n個單詞，問能否將它們連成一串，需要前面單詞的最後一個字母和後面單詞的第一個字母相同。

思路：統計每個字母出現在首尾的次數。

對於每個字母，只有可能：①首尾次數相同；②首位次數=尾巴次數+1（這個字母作爲頭）；③尾巴次數=首位次數+1（這個字母作爲尾）；④|首位次數-尾巴次數|>=2（不可能連成串）

之後統計可以當做頭和尾的字母數量。如果都爲0或都爲1，則進行dfs判斷是否整個連通；否則壓根不可能連成串。

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 int N = 410;
const double _e = 10e-6;
const int maxn = 2e5;
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, t, n, m;
int head, back;
int start[27], endd[27], has[27];
vector<int> a[27];
bool used[27];
int maze[27][27];
char s[1010];

void dfs(int u)
{
	used[u] = true;
	for (int i = 0; i < a[u].size(); i++) {
		int v = a[u][i];
		if (!used[v])
			dfs(v);
	}
}

int main()
{
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		cl0(start); cl0(endd); cl0(used); cl0(maze); cl0(has);
		for (int i = 0; i < 26; i++)
			a[i].clear();
		head = 0; back = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%s", s);
			int len = strlen(s), l = s[0] - 'a', r = s[len - 1] - 'a';
			if (maze[l][r] == 0) {
				a[l].push_back(r);
				maze[l][r] = 1;
			}
			has[l] = 1; has[r] = 1;
			start[l]++; endd[r]++;
		}
		int st;
		for (int i = 0; i < 26; i++)
			if (has[i] == 1) {
				st = i;
				break;
			}
		bool judge = true;
		for (int i = 0; i < 26; i++) {
			if (start[i] != endd[i]) {
				if (start[i] == endd[i] + 1) {
					head++;
					st = i;
				}
				else if (endd[i] == start[i] + 1)
					back++;
				else {
					judge = false;
					break;
				}
			}
		}
		if (!judge) {
			puts("The door cannot be opened.");
			continue;
		}
		judge = false;
		if ((head == 0 && back == 0) || (head == 1 && back == 1)) {
			judge = true; dfs(st);
			for (int i = 0; i < 26; i++) {
				if (has[i] == 1 && !used[i]) {
					judge = false;
					break;
				}
			}
		}
		if (judge)
			puts("Ordering is possible.");
		else
			puts("The door cannot be opened.");
	}
	return 0;
}