Sicily 3394 Pair

最新推荐文章于 2017-12-20 22:06:30 发布

原创最新推荐文章于 2017-12-20 22:06:30 发布 · 490 阅读

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本文探讨了一道有趣的图论问题，并提供了三种不同的解决方案，包括类似拓扑排序的方法、深度优先搜索（DFS）以及匈牙利算法。通过这些方法展示了如何有效地解决特定类型的图论问题。

http://soj.me/3394

一道很有趣的题目,居然写了三种做法。可以用类似拓扑排序的思想,先把入度为一的点入队,每次pop出2个,并且把图中所有指向第二个点的边删掉.如果删掉边的点入度为一了则入队.

#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define FILL(x) memset(x, 0, sizeof(x));
#define PB push_back
const int M = 10010;

int main()
{
    int T;
    cin>>T;
    while(T--) {
        vector<int> g[M];
        int f, t;
        int nn;
        int deg[M];
        FILL(deg);
        scanf("%d", &nn);
        for (int i=0; i<nn-1; i++) {
            scanf("%d%d", &f, &t);
            g[f].PB(t);
            g[t].PB(f);
            deg[f] ++;
            deg[t] ++;

        }
        queue<int> q;
        bool vis[M];
        FILL(vis);
        for (int i=1; i<=nn; i++) {
            if (deg[i] == 1) 
                q.push(i);
        }
        while(!q.empty()) {
            int f = q.front(); q.pop();
            if (deg[f] == 1 && !vis[f]) {
                int t = g[f][0];
                vis[t] = vis[f] = 1;
                g[f].clear();
                for (int i=0; i<g[t].size(); i++) {
                    int v = g[t][i];
                    for (int j=0; j<g[v].size(); j++) {
                        int x = g[v][j];
                        if (x == t) {
                            g[v].erase(g[v].begin() + j);
                            deg[v] --;
                        }
                    }
                    if (deg[v] == 1) 
                        q.push(v);
                }
            }
        }
        bool flag = 1;
        for (int i=1; i<=nn; i++) {
            if (vis[i] != 1)
                flag = 0;
        }
        if (flag) cout<<"Yes\n";
        else cout<<"No\n";
    }
}

好吧,看错了,其实是水题. 一次dfs

#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int M = 10010;
int vis[M];
vector<int> g[M];
int dfs(int f);

int main()
{
    int T;
    cin>>T;
    while(T--) {
        for (int i=0; i<M; i++) {
            g[i].clear();
        }
        int f, t, nn;
        scanf("%d", &nn);
        memset(vis, 0, sizeof(vis));
        for (int i=0; i<nn-1; i++) {
            scanf("%d%d", &f, &t);
            g[f].push_back(t);
            g[t].push_back(f);
        }
        if (dfs(1) == 0) 
            cout<<"Yes"<<endl;
        else 
            cout<<"No"<<endl;
    }
}
int dfs(int f) {
    vis[f] = 1;
    int cnt = 0;
    int flag = 0;
    for (int i=0; i<g[f].size(); i++) {
        int t=g[f][i];
        if (!vis[t])  {
            flag ++;
            cnt += dfs(t);
        }
    }
    if (flag == 0) return 1; // leaf
    if (cnt >= 2) return 2;
    else if (cnt == 1) return 0;
    else if (cnt == 0) return 1;
}

匈牙利算法

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>

using namespace std;
#ifdef DEBUG
#define cvar(x) cerr << "<" << #x << ": " << x << ">"
#define evar(x) cvar (x) << endl
#define debug(...) printf( __VA_ARGS__) 
template<class T> void DISP(const char *s, T x, int n) {cerr << "[" << s << ": "; for (int i = 0; i < n; ++i) cerr << x[i] << " "; cerr << "]" << endl;}
#define disp(x,n) DISP(#x " to " #n, x, n)
#else
#define debug(...) 
#define cvar(...) ({})
#define evar(...) ({})
#define disp(...) ({})
#endif

#define FILL(x) memset(x, 0, sizeof(x));
#define PB push_back

template<class T> inline T cub(T a){return a*a*a;}
template <typename T> T gcd(T x, T y) {for (T t; x; t = x, x = y % x, y = t); return y; }
typedef long long int64;


const int INF = 0x3f3f3f3f;
const int M = 10010;
const int MOD = int(1e9) + 7;
const double EPS = 1E-9;
const double  PI = acos(-1.0); //M_PI;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int vis[M], mat[M];
vector<int> g[M];
bool find(int i);

int main()
{
#ifdef LOCAL 
    freopen("data.in","r",stdin);  
    //freopen("data.out", "w", stdout):
#endif
    int T;
    cin>>T;
    while(T--) {
        for (int i=0; i<M; i++) {
            g[i].clear();
        }
        int f, t;
        int nn;
        scanf("%d", &nn);
        FILL(vis);
        FILL(mat);
        for (int i=0; i<nn-1; i++) {
            scanf("%d%d", &f, &t);
            g[f].PB(t);
            g[t].PB(f);
        }
        int cnt=0;
        for (int i=1; i<=nn; i++) {
            FILL(vis);
            if (!mat[i])
            if (find(i)) cnt++;
        }
        if (nn % 2 == 0 && cnt == nn/2) {
            cout<<"Yes\n";
        }
        else cout<<"No\n";
    }
    return 0;
}
bool find(int f) {
    for (int i=0; i<g[f].size(); i++) {
        int t = g[f][i];
        if (!vis[t]) {
            vis[t] = 1;
            if (mat[t] == 0 || find(mat[t]))  {
                mat[t] = f;
                return 1;
            }

        }
    }
    return 0;
}