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 Aji (Aji = 2 if the child i likes the candy j, or else Aji = 1).
The child i feels happy while 
( Cij = 1 if the child i get the candy j, or else Cij = 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: Ai1, Ai2, Ai3, ..., AiM (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
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;
}