2015
05-23

# hdu 4322-candy-图-[解题报告]hoj

http://acm.hdu.edu.cn/showproblem.php?pid=4322

Problem Description
There are N candies and M kids, the teacher will give this N candies to the M kids. The i-th kids for the j-th candy has a preference for like[i][j], if he like the sugar, like[i][j] = 1, otherwise like[i][j] = 0. If the i-th kids get the candy which he like
he will get K glad value. If he or she do not like it. He will get only one glad value. We know that the i-th kids will feel happy if he the sum of glad values is equal or greater than B[i]. Can you tell me whether reasonable allocate this N candies, make
every kid feel happy.

Input
The Input consists of several cases .The first line contains a single integer t .the number of test cases.
For each case starts with a line containing three integers N, M, K (1<=N<=13, 1<=M<=13, 2<=K<=10)
The next line contains M numbers which is B[i](0<=B[i]<=1000). Separated by a single space.
Then there are M*N like[i][j] , if the i-th kids like the j-th sugar like[i][j]=1 ,or like[i][j]=0.

Output
If there have a reasonable allocate make every kid feel happy, output "YES", or "NO".

Sample Input
2 3 2 2 2 2 0 0 0 0 0 1 3 2 2 2 2 0 0 0 0 0 0

Sample Output
Case #1: YES Case #2: NO
Hint
Give the first and second candy to the first kid. Give the third candy to the second kid. This allocate make all kids happy.
/**
hdu 4322  最大费用最大流

like[i][j]表示第i个孩子喜欢第j个糖果（总共m个孩子，n个糖）。 如果孩子拿到他喜欢的糖果，那么他将会增加k个欢乐值；
拿到不喜欢的，增加1。 如果孩子i的欢乐值大于B[i]，那么他才是开心的。能否有一种分配方案，让所有孩子都开心。

首先声明，由于被小孩子不喜欢的糖果的对小孩产生的效力是一样的，所以我们在网络流的时候先不考虑。
1 - 源点0到1~N个糖果,容量为1,费用为0
2 - 根据like数组，like[i][j] == 1时在糖果j和人N+i之间建立有一条边，容量为1，费用为0
3*- 根据b[i]和K的值建立小孩和汇点之间的边:
如果b[i] 是 K 的倍数， 说明花费b[i] / K个喜欢的糖果可以达到b[i]，建立一条边，费用为K，容量为b[i] / K;
否则，将这条边拆为两部分，第一部分是b[i] / K的部分，第二部分根据b[i] % K的部分。（如果b[i] % k == 0，说明b[i]是k的倍数;
若b[i] % k == 1, 特殊糖果和一般糖果价值一样，没必要当做特殊糖果处理）
建好图后，求最大费用最大流（只需将费用改为负的，然后套最小费用最大流即可）.。
得出特殊糖果匹配b[i]的最大值。看剩余的普通糖果是否满足缺少的b[i]。
*/
#include <iostream>
#include <string.h>
#include <algorithm>
#include <cstdio>

using namespace std;

const int oo=1e9;//无穷大
const int maxm=1111;//边的最大数量，为原图的两倍
const int maxn=42;//点的最大数量

int node,src,dest,edge;//node节点数，src源点，dest汇点，edge边数
int n,m,k,dist[maxn],lik[maxn][maxn],Flow;

struct edgenode
{
int to;//边的指向
int flow;//边的容量
int cost;//边的费用
int next;//链表的下一条边
} edges[maxm];

void prepare(int _node,int _src,int _dest);
void addedge(int u,int v,int f,int c);
bool spfa();

inline int min(int a,int b)
{
return a<b?a:b;
}

inline void prepare(int _node,int _src,int _dest)
{
node=_node;
src=_src;
dest=_dest;
for (int i=0; i<node; i++)
{
vis[i]=false;
}
edge=0;
}

void addedge(int u,int v,int f,int c)
{
edges[edge].flow=f;
edges[edge].cost=c;
edges[edge].to=v;
edges[edge].flow=0;
edges[edge].cost=-c;
edges[edge].to=u;
}

bool spfa()
{
int i,u,v,l,r=0,tmp;
for (i=0; i<node; i++) dis[i]=oo;
dis[q[r++]=src]=0;
p[src]=p[dest]=-1;
for (l=0; l!=r; ((++l>=maxn)?l=0:1))
{
{
if (edges[i].flow&&dis[v=edges[i].to]>(tmp=dis[u]+edges[i].cost))
{
dis[v]=tmp;
p[v]=i^1;
if (vis[v]) continue;
vis[q[r++]=v]=true;
if (r>=maxn) r=0;
}
}
}
return p[dest]>=0;
}

int spfaflow()
{
int i,ret=0,delta;
while (spfa())
{
//按记录原路返回求流量

for (i=p[dest],delta=oo; i>=0; i=p[edges[i].to])
{
delta=min(delta,edges[i^1].flow);
}
for (int i=p[dest]; i>=0; i=p[edges[i].to])
{
edges[i].flow+=delta;
edges[i^1].flow-=delta;
}
ret+=delta*dis[dest];
Flow+=delta;
}
return ret;
}

int main()
{
int T,tt=0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
int sum=0;
for(int i=1; i<=m; i++)
{
scanf("%d",&dist[i]);
sum+=dist[i];
}
prepare(n+m+2,0,n+m+1);
for(int i=1; i<=n; i++)
{
}
for(int i=1; i<=m; i++)
{
for(int j=1; j<=n; j++)
{
int u;
scanf("%d",&u);
if(u==1)
{
}
}
}
for(int i=1;i<=m;i++)
{
if(dist[i]%k>1)
{
}
}
printf("Case #%d: ",++tt);
Flow=0;
int cost=spfaflow();
if(sum+cost<=n-Flow)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}