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  • HS BDC HDU

    题意:

      就是混合欧拉路径板题

    解析:

      欧拉路径加一条t_ ---> s_  的边就变成了欧拉回路,所以利用这一点,如果存在两个奇点,那么这两个奇点出度大的是s_,入度大的是t_,加一条t_ ---> s_的容量为1的边即可

    然后混合欧拉回路板题

    #include <iostream>
    #include <cstdio>
    #include <sstream>
    #include <cstring>
    #include <map>
    #include <cctype>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <bitset>
    #define rap(i, a, n) for(int i=a; i<=n; i++)
    #define rep(i, a, n) for(int i=a; i<n; i++)
    #define lap(i, a, n) for(int i=n; i>=a; i--)
    #define lep(i, a, n) for(int i=n; i>a; i--)
    #define rd(a) scanf("%d", &a)
    #define rlld(a) scanf("%lld", &a)
    #define rc(a) scanf("%c", &a)
    #define rs(a) scanf("%s", a)
    #define pd(a) printf("%d
    ", a);
    #define plld(a) printf("%lld
    ", a);
    #define pc(a) printf("%c
    ", a);
    #define ps(a) printf("%s
    ", a);
    #define MOD 2018
    #define LL long long
    #define ULL unsigned long long
    #define Pair pair<int, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define _  ios_base::sync_with_stdio(0),cin.tie(0)
    //freopen("1.txt", "r", stdin);
    using namespace std;
    const int maxn = 10010, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
    int n, m, s, t, cnt;
    int f[maxn], deg[maxn], in[maxn], out[maxn], vis[maxn];
    int d[maxn], head[maxn], cur[maxn];
    set<int> ss;
    
    int find(int x)
    {
        return f[x] == x ? x : (f[x] = find(f[x]));
    }
    
    void init()
    {
        for(int i = 0; i < maxn; i++) f[i] = i;
        mem(in, 0);
        mem(head, -1);
        mem(out, 0);
        cnt = 0;
        mem(vis, 0);
        ss.clear();
    }
    
    struct edge
    {
        int u, v, c, next;
    }Edge[maxn];
    
    void add_(int u, int v, int c)
    {
        Edge[cnt].u = u;
        Edge[cnt].v = v;
        Edge[cnt].c = c;
        Edge[cnt].next = head[u];
        head[u] = cnt++;
    }
    
    void add(int u, int v, int c)
    {
        add_(u, v, c);
        add_(v, u, 0);
    }
    
    bool bfs()
    {
        queue<int> Q;
        mem(d, 0);
        Q.push(s);
        d[s] = 1;
        while(!Q.empty())
        {
            int u = Q.front(); Q.pop();
            for(int i = head[u]; i != -1; i = Edge[i].next)
            {
                edge e = Edge[i];
                if(!d[e.v] && e.c > 0)
                {
                    d[e.v] = d[e.u] + 1;
                    Q.push(e.v);
                    if(e.v == t) return 1;
                }
            }
        }
        return d[t] != 0;
    }
    
    int dfs(int u, int cap)
    {
        int ret = 0;
        if(u == t || cap == 0)
            return cap;
        for(int &i = cur[u]; i != -1; i = Edge[i].next)
        {
            edge e = Edge[i];
            if(d[e.v] == d[u] + 1 && e.c > 0)
            {
                int V = dfs(e.v, min(cap, e.c));
                Edge[i].c -= V;
                Edge[i^1].c += V;
                ret += V;
                cap -= V;
                if(cap == 0) break;
            }
        }
        if(cap > 0) d[u] = -1;
        return ret;
    }
    
    int Dinic(int u)
    {
        int ans = 0;
        while(bfs())
        {
            memcpy(cur, head, sizeof(head));
            ans += dfs(u, INF);
        }
        return ans;
    }
    
    int main()
    {
        int T, kase = 0;
        cin >> T;
        while(T--)
        {
            string str;
            int u, v, w;
            cin >> n;
            init();
            s = 0, t = 30;
            for(int i = 1; i <= n; i++)
            {
                cin >> str >> w;
                u = str[0] - 'a' + 1, v = str[str.size() - 1] - 'a' + 1;
                vis[u] = vis[v] = 1;
                in[v]++, out[u]++;
                if(u != v && w == 0) add(u, v, 1);
                int l = find(u);
                int r = find(v);
                if(l != r) f[l] = r;
            }
            int flag = 0, m_sum = 0, s_ = INF, t_ = INF;
            for(int i = 1; i <= 26; i++)
            {
                int x = find(i);
                if(vis[x]) ss.insert(x);
                if(abs(out[i] - in[i]) & 1)
                {
                    if(out[i] > in[i]) s_ = i;
                    else t_ = i;
                    if(++flag > 2) break;
                }
                if(out[i] > in[i]) add(s, i, (out[i] - in[i]) / 2), m_sum += (out[i] - in[i]) / 2;
                else if(in[i] > out[i]) add(i, t, (in[i] - out[i]) / 2);
    
            }
            if(s_ != INF && t_ != INF) add(t_, s_,  1);
         //   cout << flag << "   " << ss.size() << endl;
            printf("Case %d: ", ++kase);
            if(flag > 2 || ss.size() > 1 || flag == 1)
            {
                cout << "Poor boy!" << endl;
                continue;
            }
            if(m_sum == Dinic(s))
                cout << "Well done!" << endl;
            else
                cout << "Poor boy!" << endl;
    
        }
    
        return 0;
    }
    自己选择的路,跪着也要走完。朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其它人怎么样,我们也能够保持自己的本色走下去。
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  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/9760305.html
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