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题意
给你两个集合X,Y,X集合有N个点,Y集合有M个点,输入一个上下界down,up,现在有K条边,输入K条边(u,v)。每选择一条边(u,v),u和v点的权值就+1,问能否通过选择一些边(每条边只能选一次)使得所有点的权值都在[down,up]之间。
思路
有源汇的上下界可行流。
建立源点s,t。
①s向X集合每一个点连边,上下界为[down,up]
②Y集合向 t 连边, 上下界为[down,up]
③题目中的边,因为边只能选一次,所以上下界为[0,1]
然后跑有源汇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; //起点,终点,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; //结点数,边数(包括反向弧),源点s,汇点t
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表,G[i][j]表示结点i的第j条边在edges数组中的序号
int d[MAXN]; //从起点到i的距离(层数差)
int cur[MAXN]; //当前弧下标
bool vis[MAXN]; //BFS分层使用
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()//构造分层网络
{
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)//沿阻塞流增广
{
if (x == t || a == 0) return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)//从上次考虑的弧
{
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增广
{
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
*/