比较简单的算法:但是当点太多需要剪枝,不然很耗时

void Floyd()
{
    for(int k=0;k<n;++k)
        for(int i=0;i<n;++i)
            for(int j=0;j<n;++j)
                    dj[i][j] = min(dj[i][j],dj[i][k]+dj[k][j]);
}

 

hdu1869

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF (1<<29)
#define N   (110)

int dj[N][N];
int n,m;


void Floyd()
{
    for(int k=0;k<n;++k)
        for(int i=0;i<n;++i)
            for(int j=0;j<n;++j)
                    dj[i][j] = min(dj[i][j],dj[i][k]+dj[k][j]);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<N;++i)
            for(int j=0;j<N;++j)
            {
                if(i != j) dj[i][j] = INF;
                else dj[i][j] = 0;
            }
        for(int i=0;i<m;++i)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            if(a != b) dj[a][b] = dj[b][a] = 1;
        }
        Floyd();
        bool flag = true;
        for(int i=0;i<n;++i)
            for(int j=0;j<n;++j)
                if(dj[i][j] > 7)
                {
                    flag =false;
                    break;
                }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}
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