对拍代码四个文件

文章描述了一个使用C++编写的程序,实现了一个用于解决最大流问题的Dinic算法,包括DFS和BFS辅助函数,以及处理输入数据和输出结果的部分。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

my.cpp

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int mod = 1e9 + 7;
int n, m;

bool check(vector<int> a, vector<int> b) {
	for (int i = 0; i < n; i++) {
		if (a[i] + b[i] >= 1 && a[i] + b[i] <= m) {
			continue;
		} else {
			return false;
		}
	}
	return true;
}

void solve() {
	cin >> n >> m;
	vector<int> a(n), b(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> b[i];
	}
	sort(a.begin(), a.end());
	sort(b.begin(), b.end(), [](int& a, int& b) {
		return a > b;
	});
	if (check(a, b)) {
		cout << "Yes\n";
	} else {
		cout << "No\n";
	}
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	cin >> T;
	while (T--) {
		solve();
	}
	return 0;
}

ac.cpp

#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
 
#include <bits/stdc++.h>
 
using namespace std;
 
#define out(x) cout << #x << '=' << (x) << endl
#define out2(x, y) cout << #x << '=' << (x) << ',' << #y << '=' << (y) << endl 
#define no do { cout << "No" << endl; return; } while(0)
#define yes do { cout << "Yes" << endl; return; } while (0)
#define outvec(a) do { for (auto &v : (a)) { cout << v << ' '; } cout << endl; } while (0)
#define lowbit(x) ((x) & -(x))
#define gcd __gcd 
#define inf 0x3f3f3f3f3f3f3f3fLL
#define infi 0x3f3f3f3f
 
using ll = long long;
using pii = pair<int, int>;
 
template<typename T> ostream & operator << (ostream &out,const set<T>&obj){out<<"set(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<*it;out<<")";return out;}
template<typename T1,typename T2> ostream & operator << (ostream &out,const map<T1,T2>&obj){out<<"map(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<it->first<<": "<<it->second;out<<")";return out;}
template<typename T1,typename T2> ostream & operator << (ostream &out,const pair<T1,T2>&obj){out<<"<"<<obj.first<<", "<<obj.second<<">";return out;}
template<typename T> ostream & operator << (ostream &out,const vector<T>&obj){out<<"vector(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<*it;out<<")";return out;}

const int MAX = (1ll << 31) - 1;

struct Edge{
    int to;
    int dis;
    int next;
} edges[600000];

int cur[10010], head[10010], edge_num = -1;
int n, m, s, t;

void addEdge2(int from, int to, int dis){
    edges[++edge_num].to = to;
    edges[edge_num].dis = dis;
    edges[edge_num].next = head[from];
    head[from] = edge_num;
}

void addEdge(int from, int to, int dis){
    addEdge2(from, to, dis), addEdge2(to, from, 0);
}

int d[10010];

int DFS(int u, int flow){
    if (u == t) return flow;
    int _flow = 0, __flow;
    for (int& c_e = cur[u]; c_e != -1; c_e = edges[c_e].next){
        int v = edges[c_e].to;
        if (d[v] == d[u] + 1 && edges[c_e].dis > 0){
            __flow = DFS(v, min(flow, edges[c_e].dis));
            flow -= __flow;
            edges[c_e].dis -= __flow;
            _flow += __flow;
            edges[c_e^1].dis += __flow;
            if (!flow)
                break;
        }
    }
    if (!_flow) d[u] = -1;
    return _flow;
}

bool BFS(){
    memset(d, -1, sizeof(d));
    queue<int> que; que.push(s);
    d[s] = 0; int u, _new;
    while (!que.empty()){
        u = que.front(), que.pop();
        for (int c_e = head[u]; c_e != -1; c_e = edges[c_e].next){
            _new = edges[c_e].to;
            if (d[_new] == -1 && edges[c_e].dis > 0){
                d[_new] = d[u] + 1;
                que.push(_new);
            }
        }
    }
    return (d[t] != -1);
}

int dinic(){
    int max_flow = 0;
    while (BFS()){
        for (int i = 1; i <= n; ++i) cur[i] = head[i];
        max_flow += DFS(s, MAX);
    }
    return max_flow;
}

void solve() {
    memset(head, -1, sizeof(head));
    edge_num = -1;
    n = m = 0;
    int x, y;
    cin >> x >> y;
    vector<int> a(x + 1), b(x + 1);
    for (int i = 1; i <= x; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= x; i++) {
        cin >> b[i];
    }
    s = ++n, t = ++n;
    for (int i = 1; i <= x; i++) {
        n++;
        addEdge(s, t + i, 1);
        for (int j = 1; j <= x; j++) {
            if (a[i] + b[j] > 0 && a[i] + b[j] <= y) {
                addEdge(t + i, t + x + j, 1);
            }
        }
        n++;
        addEdge(t + x + i, t, 1);
    } 
    
    if (dinic() == x) {
        yes;
    }
    no;
}


int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int t = 1;
	cin >> t;
    
    while (t--) {
    	solve(); 
	}
}

data.cpp

#include<bits/stdc++.h>
#define int long long
using namespace std;

int Random(int mod) {
	int ans = 2147483647;
	return ans = ans * rand() % mod;
}

signed main()
{
    struct _timeb T;
    _ftime(&T);
    srand(T.millitm);
    int t = Random(30) + 1; // 1~30
    cout << t << '\n';
    
    while (t--) {
    	int n = Random(500) + 1, m = Random(500) + 1;
    	cout << n << ' ' << m << '\n';
    	for (int i = 0; i < n; i++) {
    		cout << Random(1001) - 500 << ' ';
		}
		for (int i = 0; i < n; i++) {
    		cout << Random(1001) - 500 << ' ';
		}
		cout << '\n';
	}
	return 0;
}

duipai.cpp

#include <iostream>
#include <cstdio>
#include <windows.h>
#include <cstdlib>
#include <ctime>
using namespace std;
int main()
{
    int ok = 0;
    int n = 50;
    for (int i = 1; i <= n; ++i)
    {
        system("data.exe > in.txt");
        system("ac.exe < in.txt > ac.txt");
        double begin = clock();
        system("my.exe < in.txt > my.txt");
        double end = clock();

        double t = (end - begin);
        if (system("fc ac.txt my.txt"))
        {
            printf("测试点#%d Wrong Answer\n", i);
            return -1;
        }
        else if (t > 1000) //1秒
        {
            printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
        }
        else
        {
            printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
            ok++; //AC数量+1
        }
        printf("\n");
    }
    printf("\n");
    double res = 100.0 * ok / n;
    printf("共 %d 组测试数据,AC数据 %d 组。 得分%.1lf。", n, ok, res);
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值