2020牛客暑期多校训练营（第一场）I.1 or 2

2020牛客暑期多校训练营（第一场）I.1 or 2

题目链接

题目描述

Bobo has a graph with n vertices and m edges where the i-th edge is between the vertices $a_i$ and $b_i$. Find out whether is possible for him to choose some of the edges such that the i-th vertex is incident with exactly $d_i$ edges.

输入描述:

The input consists of several test cases terminated by end-of-file.

The first line of each test case contains two integers n and m.
The second line contains n integers $d_1,d_2,\dots ,d_n$.
The i-th of the following m lines contains two integers $a_i$ and $b_i$.

$\ast 1\le n\le 50$
$\ast 1\le m\le 100$
$\ast 1\le d_i\le 2$
$\ast 1\le a_i,b_i\le n$
$\ast a_i\ne b_i$
$\ast \{a_i,b_i\}\ne \{a_j,b_j\}$
* The number of tests does not exceed 100.

输出描述:

For each test case, print “Yes” without quotes if it is possible. Otherwise, print “No” without quotes.

示例1

输入

2 1
1 1
1 2
2 1
2 2
1 2
3 2
1 1 2
1 3
2 3

输出

Yes
No
Yes

这题就是重构图然后判断完全匹配，和 HDU 3551 一模一样，(⊙﹏⊙)，考了原题都没发现，太菜了，我还一直在想怎么用 DFS 或者 BFS 甚至用离散数学的知识试图解出它，AC代码如下：

#include 
using namespace std;
const int N=1e3+5;
deque<int>q;
bool G[N][N],g[N][N],inque[N],inblossom[N],vis[N];
int match[N],pre[N],base[N];
int n,m,t,u,v,d,id;
vector<int>V[N];
struct node{
    int u,v;
}E[N];
int Find(int u,int v){
    bool inpath[N]={false};
    while(1){
        u=base[u];
        inpath[u]=true;
        if(match[u]==-1)break;
        u=pre[match[u]];
    }
    while(1){
        v=base[v];
        if(inpath[v])return v;
        v=pre[match[v]];
    }
}

void reset(int u,int anc){
    while(u!=anc){
        int v=match[u];
        inblossom[base[u]]=1;
        inblossom[base[v]]=1;
        v=pre[v];
        if(base[v]!=anc)pre[v]=match[u];
        u=v;
    }
}

void contract(int u,int v,int n){
    int anc=Find(u,v);
    memset(inblossom,0,sizeof(inblossom));
    reset(u,anc);reset(v,anc);
    if(base[u]!=anc)pre[u]=v;
    if(base[v]!=anc)pre[v]=u;
    for(int i=1;i<=n;i++)
        if(inblossom[base[i]]){
            base[i]=anc;
            if(!inque[i]){
                q.push_back(i);
                inque[i]=1;
            }
        }
}

bool dfs(int s,int n){
    for(int i=0;i<=n;i++)pre[i]=-1,inque[i]=0,base[i]=i;
    q.clear();q.push_back(s);inque[s]=1;
    while(!q.empty()){
        int u=q.front();q.pop_front();
        for(int v=1;v<=n;v++){
            if(g[u][v]&&base[v]!=base[u]&&match[u]!=v){
                if(v==s||(match[v]!=-1&&pre[match[v]]!=-1))contract(u,v,n);
                else if(pre[v]==-1){
                    pre[v]=u;
                    if(match[v]!=-1)q.push_back(match[v]),inque[match[v]]=1;
                    else{
                        u=v;
                        while(u!=-1){
                            v=pre[u];
                            int w=match[v];
                            match[u]=v;
                            match[v]=u;
                            u=w;
                        }
                        return true;
                    }
                }
            }
        }
    }
    return false;
}

void calculate()
{
    int ans=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<=id-1;i++)
        if(match[i]==-1&&dfs(i,id-1)) ans++;
    if(ans*2==id-1) printf("Yes\n");
    else printf("No\n");
}

void init(){
    id=1;
    memset(G,0,sizeof(G));
    memset(g,0,sizeof(g));
    for(int i=1;i<=n;i++) V[i].clear();
}

void build(){
    for(int i=1;i<=m;i++){
        for(auto k:V[E[i].u]) g[k][id]=g[id][k]=1;
        id++;
        for(auto k:V[E[i].v]) g[k][id]=g[id][k]=1;
        g[id][id-1]=g[id-1][id]=1;
        id++;
    }
}

int main(){
    while(~ scanf("%d%d",&n,&m)){
        init();
        for(int i=1;i<=n;i++){
            scanf("%d",&d);
            while(d--) V[i].push_back(id++);
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            G[u][v]=G[v][u]=1;
            E[i].u=u,E[i].v=v;
        }
        build();
        calculate();
    }
}