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Rikka with Graph

题目传送：HDU - 5422 - Rikka with Graph

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m;

int main() {
    while(scanf("%d %d", &n, &m) != EOF) {
        int u, v;
        int flag = 0;
        for(int i = 0; i < m; i ++) {
            scanf("%d %d", &u, &v);
            if((u == 1 && v == n) || (u == n && v == 1)) flag = 1;
        }
        if(flag == 1) printf("1 %d\n", n * (n - 1) / 2);
        else printf("1 1\n");
    }
    return 0;
}

Rikka with Tree

题目传送：HDU - 5423 - Rikka with Tree

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 1005;
int n;
vector<int> G[maxn];

bool judge1() {
    if(G[1].size() <= 0 || G[1].size() >= 2) {
        return false;
    }
    int pre = 1;
    int v = G[1][0];
    while(G[v].size() == 2) {
        for(int i = 0; i < 2; i ++) {
            if(G[v][i] != pre) {
                pre = v;
                v = G[v][i];
                if(G[v].size() == 1) return true;
                break;
            }
        }
    }
    return false;
}

bool judge2() {
    if(G[1].size() != n - 1) return false;
    return true;
}

bool judge3() {
    if(G[1].size() != 1) {
        return false;
    }
    int cnt = 1;
    int pre = 1;
    int v = G[1][0];
    while(G[v].size() == 2) {
        for(int i = 0; i < 2; i ++) {
            if(G[v][i] != pre) {
                pre = v;
                v = G[v][i];
                cnt ++;
                break;
            }
        }
        if(G[v].size() != 2) break;
    }
    //cout << n << " " << cnt << endl;
    if(G[v].size() == n - cnt) return true;
    return false;
}

int main() {
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i < maxn; i ++) G[i].clear();

        int u, v;
        for(int i = 1; i < n; i ++) {
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }

        if(judge1() || judge2() || judge3()) {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}

Rikka with Graph II

题目传送：HDU - 5424 - Rikka with Graph II

官方题解：

如果图是连通的，可以发现如果存在哈密顿路径，一定有一条哈密顿路径的一端是度数最小的点，从那个点开始直接DFS搜索哈密顿路径复杂度是O(n)的。要注意先判掉图不连通的情况。

AC代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 1005;
int n;

vector<int> G[maxn];
map<pair<int, int>, int> mp;
int deg[maxn];
int vis[maxn];

bool dfs(int u, int cnt) {
    vis[u] = 1;
    if(cnt == n) return true;
    int d = G[u].size();
    for(int i = 0; i < d; i ++) {
        int v = G[u][i];
        if(!vis[v]) {
            if(dfs(v, cnt + 1)) return true;
            vis[v] = 0;
        }
    }
    return false;
}

int cnt;
void judge(int u) {
    vis[u] = 1;
    cnt ++;
    int d = G[u].size();
    for(int i = 0; i < d; i ++) {
        int v = G[u][i];
        if(!vis[v]) {
            judge(v);
        }
    }
}

int main() {
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i <= n; i ++) G[i].clear();
        mp.clear();
        memset(deg, 0, sizeof(deg));
        int u, v;
        for(int i = 0; i < n; i ++) {
            scanf("%d %d", &u, &v);
            if(u == v) continue;
            if(mp.find(make_pair(u, v)) != mp.end() || mp.find(make_pair(v, u)) != mp.end() ) {
                continue;
            }
            G[u].push_back(v);
            G[v].push_back(u);
            deg[u] ++;
            deg[v] ++;
            mp[make_pair(u, v)] = 1;
        }

        int mi = INF;
        int id;
        for(int i = 1; i <= n; i ++) {
            if(deg[i] < mi) {
                mi = deg[i];
                id = i;
            }
        }

        memset(vis, 0, sizeof(vis));
        cnt = 0;
        judge(1);//特判不连通的情况，因为如果不连通直接dfs会因为回溯太多次而超时
        if(cnt != n) {
            printf("NO\n");
            continue;
        }

        memset(vis, 0, sizeof(vis));
        //cout<< id << endl;
        if(dfs(id, 1)) {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}

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