HDU 3605 Escape（最大流）

Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s): 691

Problem Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

 

 

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

 

 

Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

 

 

Sample Input

1 1 1 1 2 2 1 0 1 0 1 1

 

 

Sample Output

YES NO

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

 

 

Recommend

lcy

 

 

 

这题一看题目是很简单的最大流。

但是数据好大，试了很多模板，都是TLE。

 

后来发现可以合并点，

因为m<=10.

所以用二进制记录。

在n个点中，如果是一样的就合并。

这样最多是1024+m+2个点。

 

 

但是这题还是很坑。。。。在HDU上用G++交无论如何都是TLE的。直接读入数据输出都是TLE.

改成C++就AC了。。。

 

 

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;

const int MAXN=100110;
const int MAXM=4000110;
const int INF=0x3f3f3f3f;
struct Node
{
    int to,next,cap;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++;
    edge[tol].to=u;edge[tol].cap=rw;edge[tol].next=head[v];head[v]=tol++;
}
int sap(int start,int end,int nodenum)
{
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    memcpy(cur,head,sizeof(head));
    int u=pre[start]=start,maxflow=0,aug=-1;
    gap[0]=nodenum;
    while(dis[start]<nodenum)
    {
        loop:
        for(int  &i=cur[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap&&dis[u]==dis[v]+1)
            {
                if(aug==-1||aug>edge[i].cap)
                    aug=edge[i].cap;
                pre[v]=u;
                u=v;
                if(v==end)
                {
                    maxflow+=aug;
                    for(u=pre[u];v!=start;v=u,u=pre[u])
                    {
                        edge[cur[u]].cap-=aug;
                        edge[cur[u]^1].cap+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        }
        int mindis=nodenum;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap&&mindis>dis[v])
            {
                cur[u]=i;
                mindis=dis[v];
            }
        }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}

int num[1025];
int a[11];
int bit[11];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    bit[0]=1;
    for(int i=1;i<=10;i++)bit[i]=bit[i-1]*2;
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
        {
            int tmp=0;
            for(int j=0;j<m;j++){scanf("%d",&a[j]);tmp+=a[j]*bit[j];}
            num[tmp]++;
        }
        int start=0,end=1024+m+1,nodenum=2014+m+2;
        for(int i=0;i<1024;i++)
        {
            if(num[i]==0)continue;
            addedge(start,i+1,num[i]);
            for(int j=0;j<10;j++)
                if(i&bit[j] )
                    addedge(i+1,1024+j+1,INF);
        }

        int tmp;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&tmp);
            addedge(i+1024,end,tmp);
        }

        if(sap(start,end,nodenum)==n)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}