題意

給你兩個集合X,Y，X集合有N個點，Y集合有M個點，輸入一個上下界down,up，現在有K條邊，輸入K條邊（u，v）。每選擇一條邊(u，v)，u和v點的權值就+1，問能否通過選擇一些邊(每條邊只能選一次)使得所有點的權值都在[down,up]之間。

思路

有源匯的上下界可行流。

建立源點s，t。

①s向X集合每一個點連邊，上下界為[down,up]

②Y集合向 t 連邊， 上下界為[down,up]

③題目中的邊，上下界為[0,1]

然後跑有源匯的上下界可行流即可。

//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, flow;       //起點,終點,容量,流量
    Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct Dinic
{
    int n, m, s, t;                //結點數,邊數(包括反向弧),源點s,匯點t
    vector<Edge> edges;            //邊表。edges[e]和edges[e^1]互為反向弧
    vector<int> G[MAXN];           //鄰接表，G[i][j]表示結點i的第j條邊在edges陣列中的序號
    int d[MAXN];                   //從起點到i的距離（層數差）
    int cur[MAXN];                 //當前弧下標
    bool vis[MAXN];                //BFS分層使用

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void AddEdge(int from, int to, int cap)
    {
        edges.push_back(Edge(from, to, cap, 0));
        edges.push_back(Edge(to, from, 0, 0));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BFS()//構造分層網路
    {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        d[s] = 0;
        vis[s] = true;
        Q.push(s);
        while (!Q.empty())
        {
            int x = Q.front(); Q.pop();
            for (int i = 0; i < G[x].size(); i++)
            {
                Edge& e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow)
                {
                    vis[e.to] = true;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x, int a)//沿阻塞流增廣
    {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int& i = cur[x]; i < G[x].size(); i++)//從上次考慮的弧
        {
            Edge& e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增廣
            {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }
    int MaxFlow(int s, int t)
    {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS())
        {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }

}solve;

int n, m, k, down, up, sum[MAXN];

int main()
{
    int CASE = 1;
    while (~scanf("%d%d%d", &n, &m, &k))
    {
        int tot = 0;
        memset(sum, 0, sizeof(sum));
        scanf("%d%d", &down, &up);
        int s = 0, t = n+m+1, vs = n+m+2, vt = n+m+3;
        solve.init(vt);
        solve.AddEdge(t, s, INF);
        for (int i = 1; i <= n; i++)
        {
            solve.AddEdge(s, i, up-down);
            sum[s] -= down;
            sum[i] += down;
        }
        for (int i = 1; i <= m; i++)
        {
            solve.AddEdge(i+n, t, up-down);
            sum[i+n] -= down;
            sum[t] += down;
        }
        while (k--)
        {
            int u, v; scanf("%d%d", &u, &v);
            solve.AddEdge(u, v+n, 1);
        }
        for (int i = s; i <= t; i++)
        {
            if (sum[i] < 0) solve.AddEdge(i, vt, -sum[i]);
            else solve.AddEdge(vs, i, sum[i]), tot += sum[i];
        }
        printf("Case %d: ", CASE++);
        int ans = solve.MaxFlow(vs, vt);
        if (ans == tot) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

/*
3 3 7
2 3
1 2
2 3
1 3
3 2
3 3
2 1
2 1
3 3 7
3 4
1 2
2 3
1 3
3 2
3 3
2 1
2 1
*/