2020牛客多校第一場 I 1 or 2

1 or 2

• 題意
• 思路
• 代碼

連結

題意

給定n個點m條邊的圖，和n個di值，對于1<=i<=n，問能否删除某些邊使得，第i個點所連邊數都為di

思路

當時這題旁邊有人用最大流過了，但是真算法是一般圖最大比對。

需要把所有點進行以下處理。

大佬部落格

這個拆度和拆邊思路可以記一下

雖然的确不知道原理。

要注意本題d的範圍為1或2。

拆邊拆度之後所有的d都為1，即一般圖比對。

代碼

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 310, maxM = 4e3+ 7;
int N, M, head[maxN], cnt, L, id[55][2], d[maxN];
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
} edge[maxM];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
struct DHS
{
    int match[maxN], pre[maxN], vis[maxN], fa[maxN], tim[maxN], idx, ans, node;
    queue<int> Q;
    inline int fid(int x) { return x == fa[x] ? x : fa[x] = fid(fa[x]); }
    inline int lca(int x, int y)
    {
        for(++idx; ; swap(x, y))
        {
            if(x)
            {
                x = fid(x);
                if(tim[x] == idx) return x;
                else
                {
                    tim[x] = idx;
                    x = pre[match[x]];
                }
            }
        }
    }
    void blossom(int x, int y, int p)
    {
        while (fid(x) != p)
        {
            pre[x] = y; y = match[x];
            if (vis[y] == 2) { vis[y] = 1; Q.push(y); }
            if (fid(x) == x) fa[x] = p;
            if (fid(y) == y) fa[y] = p;
            x = pre[y];
        }
    }
    int Aug(int S)
    {
        for(int i=1; i<=node; i++)
        {
            vis[i] = pre[i] = 0;
            fa[i] = i;
        }
        while (!Q.empty()) Q.pop();
        Q.push(S);vis[S]=1;
        while (!Q.empty())
        {
            int u = Q.front(); Q.pop();
            for (int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if (fid(u) == fid(v) || vis[v] == 2) continue;
                if (!vis[v])
                {
                    vis[v] = 2; pre[v] = u;
                    if (!match[v])
                    {
                        for (int x = v, lst; x; x = lst)
                        {
                            lst = match[pre[x]];
                            match[x] = pre[x];
                            match[pre[x]] = x;
                        }
                        return 1;
                    }
                    vis[match[v]] = 1;
                    Q.push(match[v]);
                }
                else
                {
                    int gg = lca(u, v);
                    blossom(u, v, gg);
                    blossom(v, u, gg);
                }
            }
        }
        return 0;
    }
    inline int solve()
    {
        for(int i=1; i<=node; i++)
        {
            if(!match[i]) ans += Aug(i);
        }
        return ans;
    }
} dhs;
inline void init()
{
    cnt = 0; dhs.ans = 0; dhs.idx = 0;
    for(int i=1; i<=dhs.node; i++) { head[i] = -1; dhs.match[i] = dhs.tim[i] = 0; }
}
int main()
{
    while(scanf("%d%d", &N, &M) != EOF)
    {
        dhs.node = 0;
        int sum = 0;
        for(int i=1; i<=N; i++)
        {
            scanf("%d", &d[i]); sum += d[i];
            for(int j=0; j<d[i]; j++) id[i][j] = ++dhs.node;
        }
        init();
        for(int i=1, u, v; i<=M; i++)
        {
            scanf("%d%d", &u, &v);
            dhs.node++;
            dhs.match[dhs.node] = dhs.tim[dhs.node] = 0;
            head[dhs.node] = -1;
            for(int i=0;i<d[u];i++){
            	_add(id[u][i],dhs.node);
			}
			
			dhs.node++;
            dhs.match[dhs.node] = dhs.tim[dhs.node] = 0;
            head[dhs.node] = -1;
            for(int i=0;i<d[v];i++){
            	_add(id[v][i],dhs.node);
			}
			_add(dhs.node,dhs.node-1);
           /* if(d[u]  == 2 && d[v] == 2)
            {
                dhs.node++;
                dhs.match[dhs.node] = dhs.tim[dhs.node] = 0;
                head[dhs.node] = -1;
                _add(id[u][0], dhs.node);
                _add(id[u][1], dhs.node);
                dhs.node++;
                dhs.match[dhs.node] = dhs.tim[dhs.node] = 0;
                head[dhs.node] = -1;
                _add(dhs.node - 1, dhs.node);
                _add(id[v][0], dhs.node);
                _add(id[v][1], dhs.node);
                sum += 2;
            }
            else
            {
                for(int j=0; j<d[u]; j++)
                {
                    for(int k=0; k<d[v]; k++)
                    {
                        _add(id[u][j], id[v][k]);
                    }
                }
            }*/
        }
        printf(dhs.solve() *2 == dhs.node ? "Yes\n" : "No\n");
      //  cout<<dhs.node<<endl; 
    }
    return 0;
}