Starry

发表于 更新于 分类于 本文字数： 554 阅读时长 ≈ 2 分钟

二分答案

最近失智了

更：我是zz，写出了模拟怀疑复杂度没冲

模拟：

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#include <algorithm>
#include <bitset>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <cmath>
#include <stack>
#include <set>
#include <queue>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))

#define fi first
#define se second
#define sz(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;

LL tt, n, m;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> tt;
    f(sb, 1, tt)
    {
        cin >> n >> m;
        vector v(n, 0);
        for (int i = 1, x; i <= m; ++i)
        {
            cin >> x;
            v[x - 1] += 1;
        }
        sort(all(v));
        multiset<int> s;
        for (auto it : v)
            s.insert(it);
        while (*(prev(s.end())) - *s.begin() >= 3)
        {
            auto mi = *s.begin(), ma=*(prev(s.end()));
            s.erase(s.begin());
            s.erase(prev(s.end()));
            ma-=1,mi+=2;
            s.insert(ma);
            s.insert(mi);
        }
        cout<<*(prev(s.end()))<<"\n";
    }
    return 0;
}

二分：

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#include <algorithm>
#include <bitset>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <cmath>
#include <stack>
#include <set>
#include <queue>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))

#define fi first
#define se second
#define sz(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;

LL tt, n, m;

struct Node
{
    int a,b;
    bool operator<(const Node &bb){
        return a+b*2<bb.a+bb.b*2;
    }
};


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> tt;
    f(sb, 1, tt)
    {
        cin >> n >> m;
        vector v(n, 0);
        for (int i = 1, x; i <= m; ++i)
        {
            cin >> x;
            v[x - 1] += 1;
        }
        LL l=1,r=m;
        auto ck=[&](int x){
            LL res=0;
            for(auto &it:v){
                if(it>x)res+=it-x;
                else res-=(x-it)/2;
            }
            return res<=0;
        };

        while(l<r){
            int mid=(l+r)>>1;
            if(ck(mid))r=mid;
            else l=mid+1;
        }
        cout<<r<<"\n";
    }
    return 0;
}

发表于 更新于 分类于 本文字数： 279 阅读时长 ≈ 1 分钟

转化为最小可相交路径覆盖问题

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#include <cstdio>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;

int n, m, ans, a[21][21], mp[21][21][21][21];
char c;
bitset<404> vis;

int main()
{
    scanf("%d%d", &n, &m);
    vector<vector<int>> g(n * m + 1);
    vector<int> f(n * m + 1, 0);

    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= m; ++j)
        {
            scanf(" %c", &c);
            ans += (c - '0');
            a[i][j] = c - '0';
            if(a[i][j]){
                if(a[i-1][j])mp[i-1][j][i][j]=1;
                if(a[i][j-1])mp[i][j-1][i][j]=1;
            }
        }
    }

    for(int kk=1;kk<=n;++kk){
        for(int k=1;k<=m;++k){
            for(int ii=1;ii<=n;++ii){
                for(int i=1;i<=m;++i){
                    for(int jj=1;jj<=n;++jj){
                        for(int j=1;j<=m;++j){
                            if(ii!=jj&&i!=j){
                                if(mp[ii][i][kk][k]&&mp[kk][k][jj][j]){
                                    mp[ii][i][jj][j]=1;
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    for(int i=1;i<=n;++i){
        for(int ii=1;ii<=m;++ii){
            for(int j=1;j<=n;++j){
                for(int jj=1;jj<=m;++jj){
                    if(mp[i][ii][j][jj]){
                        g[(i-1)*m+ii].push_back((j-1)*m+jj);
                    }
                }
            }
        }
    }

    function<bool(int)> ck = [&](int x) -> bool
    {
        for (vector<int>::iterator it = g[x].begin(); it != g[x].end(); ++it)
        {
            if (!vis[*it])
            {
                vis[*it] = 1;
                if (!f[*it] || ck( f[*it]))
                {
                    f[*it] = x;
                    return true;
                }
            }
        }
        return false;
    };
    for (int i = 1; i <= n*m; ++i)
    {
        vis.reset();
        if (ck(i))
        {
            ans -= 1;
        }
    }

    printf("%d", ans);
}

发表于 更新于 分类于 本文字数： 200 阅读时长 ≈ 1 分钟

最小不相交路径覆盖问题，最小路径覆盖=原图结点数-新图匹配数

路径输出开个bitset记录每个点是否是存在后继，然后递归输出即可。

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#include <cstdio>
#include <vector>
#include <bitset>
#include<iostream>
using namespace std;

int n, m,ans;
bitset<404> vis,st;


int main()
{
    scanf("%d%d", &n, &m);
    vector<vector<int>> g(n + 1);
    vector<int> f(n + 1, 0);

    for (int j = 1, x, y; j <= m; ++j)
    {
        scanf("%d%d", &x, &y);
        g[x].push_back(y);
    }
    auto ck = [&](auto self, int x) -> bool
    {
        for (vector<int>::iterator it = g[x].begin(); it != g[x].end(); ++it)
        {
            if (!vis[*it])
            {
                vis[*it] = 1;
                if (!f[*it] || self(self, f[*it]))
                {
                    st[x]=1;
                    f[*it] = x;
                    return true;
                }
            }
        }
        return false;
    };
    for (int i = 1; i <= n; ++i)
    {
        vis.reset();
        if (ck(ck, i))
        {
            ans += 1;
        }
    }

    auto write=[&](int x,auto self)->void{
        if(f[x]){
            self(f[x],self);
            printf(" %d",x);
        }else printf("%d",x);
    };

    for(int i=1;i<=n;++i){
        if(!st[i]){
            write(i,write);
            printf("\n");
        }
    }
    printf("%d", n-ans);
}

发表于 更新于 分类于 本文字数： 108 阅读时长 ≈ 1 分钟 
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#include<cstdio>
#include<bitset>
#include<vector>

using namespace std;

int k,n,m;
bitset<505> vis;

int main(){
 while(~scanf("%d",&k)){
    if(!k)return 0;
    int ans=0;
    scanf("%d%d",&n,&m);
    vector<vector<int>> g(n+1);
    vector<int> f(m+1);
    for(int i=1,x,y;i<=k;++i)scanf("%d%d",&x,&y),g[x].push_back(y);
    auto ck=[&](auto self,int x)->bool{
        for(auto it:g[x]){
            if(!vis[it]){
                vis[it]=1;
                if(!f[it]||self(self,f[it])){
                    f[it]=x;
                    return true;
                }
            }
        }
        return false;
    };
    for(int i=1;i<=n;++i){
        vis.reset();
        if(ck(ck,i))ans+=1;
    }
    printf("%d\n",ans);
 }
}

发表于 更新于 分类于 本文字数： 4.8k 阅读时长 ≈ 18 分钟

Wordpress的编辑器对大文本的支持也太差了，，，放这备份

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