Educational Codeforces Round 8 F. Bear and Fair Set（最大流 | Hall定理）★

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图论—网络流

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本文介绍了一种利用网络流算法解决特定集合构建问题的方法。针对给定的元素范围和数量限制，通过构建合适的网络流模型来判断是否能够找到一个满足条件的集合，即该集合中模5余数为0到4的元素数量相等，并且满足额外的限制条件。

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题意:

给定N,B,Q≤104,N能被5整除
N为set大小(无相同元素),元素范围为[1,B],需满足set里元素模5的余数为[0,4]的元素个数相等
Q个条件,bi cnti,表示[1,bi]应该有cnti个数
问这种set是否存在,存在输出“fair”,否则“unfair”

分析:

把题目条件(B,N)和Q次询问一起按照Bi排序,这样形成了Q+1个不相交区间(加上一个(0,0))
建图,源点→Remainders0∼4,容量为n/5;区间→汇点,容量为cnti
Remaindersi→区间,容量为区间里数模5余Remainderi的个数
考虑求Remainders的匹配数,存在性即可以最大流判断是不是N

网络流求解，666

代码:

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 3100
#define MOD 1000000007

const int MAXN = 100010;
const int MAXM = 400010;
struct Edge
{
    int to,next,cap,flow;
} edge[MAXM];
int tot,n,m,q;
int head[MAXN];
int gap[MAXN],dep[MAXN];
int pre[MAXN],cur[MAXN];

void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
void add_edge(int u,int v,int w,int rw = 0)
{
    edge[tot].to = v;
    edge[tot].cap = w;
    edge[tot].flow = 0;
    edge[tot].next = head[u];
    head[u] = tot++;
    edge[tot].to = u;
    edge[tot].cap = rw;
    edge[tot].flow = 0;
    edge[tot].next = head[v];
    head[v] = tot++;
}
int sap(int start,int en,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == en)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        int flag = 0;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = 1;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])
            return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)
            u = edge[pre[u]^1].to;
    }
    return ans;
}
struct node
{
    int x,y;
    bool operator<(const node&a)const
    {
        return x < a.x;
    }
}a[10010];
int solve(int x,int y,int k)
{
    int k1 = x % 5;
    int k2 = y % 5;
    x += 5 - k1;
    y -= k2 + 1;
    return (y - x + 1) / 5 + (k1 <= k) + (k2 >= k);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t,C = 1;
    //scanf("%d",&t);
    while(scanf("%d%d%d",&n,&m,&q) != EOF)
    {
        init();
        int flag = 1;
        for(int i = 1; i <= q; i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        sort(a+1,a+q+1);
        a[0].x = 0;
        a[0].y = 0;
        for(int i = 1; i <= q; i++)
        {
            int k = a[i].y-a[i-1].y;
            add_edge(0,i,k);
            if(k < 0 || k > a[i].x - a[i-1].x)
                flag = 0;
            for(int j = 1; j <= 5; j++)
                add_edge(i,q+1+j,solve(a[i-1].x+1,a[i].x,j-1));
        }
        if(a[q].x != m)
        {
            int k = n-a[q].y;
            add_edge(0,q+1,k);
            if(k < 0 || k > m - a[q].x)
                flag = 0;
            for(int j = 1; j <= 5; j++)
                add_edge(q+1,q+1+j,solve(a[q].x+1,m,j-1));
        }
        for(int i = 1; i <= 5; i++)
            add_edge(q+1+i,q+7,n/5);
        if(flag && sap(0,q+7,q+8) == n)
            printf("fair\n");
        else
            printf("unfair\n");

    }
    return 0;
}