joj 2453 candy 网络流建图的题

Problem D: Candy

As a teacher of a kindergarten, you have many things to do during a day, one of which is to allot candies to all children in your class. Today you have N candies for the coming M children. Each child likes different candy, and as a teacher who know them well, you can describe how the child i likes the candy j with a number A_ji (A_ji = 2 if the child i likes the candy j, or else A_ji = 1).

The child i feels happy while ( C_ij = 1 if the child i get the candy j, or else C_ij = 0). Now your task is to allot the candies in such a way that makes every child happy (of course except you, ^_^).

Input

The first line of the input contains a single integer T (1 <= T <= 10), representing the number of cases that follow.

The first line of each case consists of two integers N and M (1 <= N <= 100000, 1 <= M <= 10), which are the number of candies and the number of children.

There are N lines following, the ith line containing M integers: A_i1, A_i2, A_i3, ..., A_iM (1 <= Aij <= 2)

The last line of the case consists of M integers: B1, B2, B3, ..., BM (0 <= Bi <= 1000000000).

Output

For each case, if there is a way to make all children happy, display the word “Yes”. Otherwise, display the word “No”.

Sample Input

2
4 3
1 2 1
2 1 1
1 1 2
1 2 2
3 2 2
1 1
1
2

Sample Output

Yes
No

网络流，主要是建图

分配的时候肯定会优先给每个孩子分配喜欢的糖果，所以先只考虑Aij=2的孩子和糖果(i,j)。

如果Ai,j=2，那么把孩子i向糖果j连一条容量为1的边，再建立源点S，向每个孩子连一条容量为Bi/2的边（因为每个开心值为2的糖果只算1，所以孩子的B值也要先除以2），最后把每个糖果向汇点T连容量为1的边，做一次网络最大流。

假设S到孩子i的流量为fi，说明孩子i已经获得了fi*2点快乐值，还需要Bi-fi*2点，这时候f1+f2+..+fm是总共分出去的糖果数，那么还剩N-(f1+f2+..+fm)个糖果，如果这个数>=sigma(Bi-fi*2)，即剩余的糖果数大于等于孩子还需要的总共快乐值，则有解，否则无解

PS：每个孩子平均能吃10000个糖，我真是无限ORZ

以下使用的是刘汝佳白书上的DINIC算法模板做的

#include<iostream>
#include<algorithm>
#include<cstring>
#define size_num 100200
#include<vector>
#include<queue>
#define INF 1e8
using namespace std;
int child[105];
struct Dinic
{
	struct Edge{int from,to,cap,flow;};
	vector<Edge> edges;
	//边表。edges[e]和edges[e+1]互为反向弧,
	//注意到e必须是偶数即是大的奇数与比他小的偶数互为反向边，即e与e^1互为反向边
	vector<int> G[size_num];
	//领接表，G[i][j]表示节点i的第j条边在e数组中的序号
	void add_edge(int from,int to,int cap)
	{
		edges.push_back((Edge){from,to,cap,0});//加入正向边
		edges.push_back((Edge){to,from,0,0});//加入反向边
		int m=edges.size();
		G[from].push_back(m-2);//存的是边的位子
		G[to].push_back(m-1);//貌似有一种静态链表的感觉
	}
	int s,t;//源点编号和汇点编号
	bool vis[size_num];//bfs时使用
	int d[size_num];//从起点到i的距离
	int cur[size_num];//当前弧的下标
	void init()
	{
		edges.clear();
		for(int i=0;i<size_num;i++)
		G[i].clear();
	}
	bool bfs()
	{
		memset(vis,0,sizeof(vis));
		queue<int > q;
		q.push(s);
		d[s]=0;
		vis[s]=1;
		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]=1;
					d[e.to]=d[x]+1;
					q.push(e.to);
				}

			}
		}
		return vis[t];
	}

	//dfs
	int dfs(int x,int a)
	{
		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)
			{
				e.flow+=f;//增加正向的流量
				edges[G[x][i]^1].flow-=f;//减少反向的流量
				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;
void read()
{
	solve.init();
	int n,m;//糖果数量和孩子的数量
	cin>>n>>m;
	int s=0,t=1+m+n;
	//solve->n=t+1;
	//1->m表示孩子，m+1->m+n表示糖果
	for(int i=1;i<=n;i++)
	{
		solve.add_edge(i+m,t,1);
		for(int j=1;j<=m;j++)
		{
			int temp;
			cin>>temp;
			if(temp==2)
				solve.add_edge(j,m+i,1);
		}
	}
	long long sum=0;
	for(int i=1;i<=m;i++)
	{
		cin>>child[i];
		sum+=child[i];
		solve.add_edge(s,i,child[i]/2);
	}
	int f=solve.maxflow(s,t);
	int yu=n-f;
	if(sum<=yu+f*2)
		cout<<"Yes\n";
	else
		cout<<"No\n";
}

int main()
{

	int T;cin>>T;
	while(T--)
	read();
	return 0;
}

以下是不用vector的代码比较快0.5s上一个是3.07秒

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

const int maxn = 100055;
const int maxm = 600005;
const int inf = 0x3f3f3f3f;
struct MaxFlow
{
	int net[maxn], gap[maxn], dis[maxn], pre[maxn], cur[maxn];
	int siz, n;
	std::queue <int> Q;
	struct EDGE
	{
		int v, cap, next;
		EDGE(){}
		EDGE(int a, int b, int c): v(a), cap(b), next(c){}
	}E[maxm<<1];
	void init(int _n)//要传入节点数
	{
		n = _n, siz = 0;
		memset(net, -1, sizeof(net));
	}
	void add_edge(int u, int v, int cap)//加边操作
	{
		E[siz] = EDGE(v, cap, net[u]);
		net[u] = siz++;
		E[siz] = EDGE(u, 0, net[v]);
		net[v] = siz++;
	}
	void bfs(int st)//广搜
	{
		int u, v;
		for(int i = 0; i <= n; i++)
			dis[i] = n, gap[i] = 0;
		gap[0] = 1, dis[st] = 0;
		Q.push(st);
		while(!Q.empty())
		{
			u = Q.front();
			Q.pop();
			for(int i = net[u]; i != -1; i = E[i].next)
			{
				v = E[i].v;
				if(!E[i^1].cap || dis[v] < n)
					continue;
				dis[v] = dis[u] + 1;
				gap[dis[v]]++;
				Q.push(v);
			}
		}
	}
	int isap(int st, int en)//st 是源点 en 是汇点
	{
		int u = pre[st] = st, ma = 0, aug = inf, v;
		bfs(en);
		for(int i = 0; i <= n; i++)
			cur[i] = net[i];
		while(dis[st] <= n)
		{
loop:		for(int &i = cur[u]; v = E[i].v, i != -1; i = E[i].next)
				if(E[i].cap && dis[u] == dis[v] + 1)
				{
					aug = std::min(aug, E[i].cap);
					pre[v] = u, u = v;
					if(v == en)
					{
						ma += aug;
						for(u = pre[u]; v != st; v = u, u = pre[u])
						{
							E[cur[u]].cap -= aug;
							E[cur[u]^1].cap += aug;
						}
						aug = inf;
					}
					goto loop;
				}
			int mi = n;
			for(int i = net[u]; v = E[i].v, i != -1; i = E[i].next)
				if(E[i].cap && mi > dis[v])
				{
					cur[u] = i;
					mi = dis[v];
				}
			if(--gap[dis[u]] == 0)
				break;
			gap[dis[u]=mi+1]++;
			u = pre[u];
		}
		return ma;
	}
};

MaxFlow G;

int main()
{
	int t, n, m, st, en, temp;
	long long sum;

	scanf("%d", &t);
	while(t--)
	{
		scanf("%d %d", &n, &m);
		st = 0, en = n + m + 1;
		G.init(en);
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= m; j++)
			{
				scanf("%d", &temp);
				if(temp == 2)
					G.add_edge(i, j+n, 1);
			}
		sum = 0;
		for(int i = 1; i <= m; i++)
		{
			scanf("%d", &temp);
			sum += temp;
		}
		for(int i = 1; i <= n; i++)
			G.add_edge(st, i, 1);
		for(int i = 1; i <= m; i++)
			G.add_edge(i+n, en, inf);
		if(((long long)n+G.isap(st, en)) >= sum)
			puts("Yes");
		else
			puts("No");
	}
	return 0;
}