HDU 3605 Escape(多重KM，4级)

最新推荐文章于 2017-11-17 16:38:00 发布

剑不飞

最新推荐文章于 2017-11-17 16:38:00 发布

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本文介绍了一个关于如何确定人类是否能在特定星球上生存的问题，并通过多重KM匹配算法来解决这一挑战。输入包括地球上的人数、可用的星球数量、每个人能适应居住的星球以及每个星球能容纳的最大人数。

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F - Escape

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:

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

Sample Output

思路：多重KM，需要理解KM。建容量cap

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof f);
using namespace std;
const int mpeo=1e5+9;
const int mxx=15;
int num[mxx],cap[mxx];
int X[mxx][mpeo];
int g[mpeo][mxx];
bool T[mxx];
int n,m;
bool dfs(int u)
{
  FOR(i,1,m)
  if(!T[i]&&g[u][i])
  { T[i]=1;
    if(num[i]<cap[i])
    {
      X[i][++num[i]]=u;
      return 1;
    }
    else
    {
      FOR(j,1,cap[i])
      {
        if(dfs(X[i][j]))
        {
          X[i][j]=u;return 1;
        }
      }
    }
  }
  return 0;
}
bool getans()
{
  clr(X,-1);clr(num,0);
  FOR(i,1,n)
  {
    clr(T,0);
    if(!dfs(i))return 0;
  }
  return 1;
}
int main()
{ int a;
 while(~scanf("%d%d",&n,&m))
 {
   FOR(i,1,n)FOR(j,1,m)
   {scanf("%d",&g[i][j]);
   }
   FOR(j,1,m)scanf("%d",&cap[j]);
   if(getans())puts("YES");
   else puts("NO");
 }
}