BestCoder Round #53

来源：互联网发布：听音乐的软件编辑：程序博客网时间：2024/05/18 02:40

现在博客更新比较少了，就当我还活着吧

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 0x7fffffffusing 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 0x7fffffffusing 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 0x7fffffffusing 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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