Kingdom
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 142 Accepted Submission(s): 84
Special Judge
Problem Description
Teacher Mai has a kingdom consisting of n cities. He has planned the transportation of the kingdom. Every pair of cities has exactly a one-way road.
He wants develop this kingdom from one city to one city.
Teacher Mai now is considering developing the city w. And he hopes that for every city u he has developed, there is a one-way road from u to w, or there are two one-way roads from u to v, and from v to w, where city v has been developed before.
He gives you the map of the kingdom. Hope you can give a proper order to develop this kingdom.
He wants develop this kingdom from one city to one city.
Teacher Mai now is considering developing the city w. And he hopes that for every city u he has developed, there is a one-way road from u to w, or there are two one-way roads from u to v, and from v to w, where city v has been developed before.
He gives you the map of the kingdom. Hope you can give a proper order to develop this kingdom.
Input
There are multiple test cases, terminated by a line "0".
For each test case, the first line contains an integer n (1<=n<=500).
The following are n lines, the i-th line contains a string consisting of n characters. If the j-th characters is 1, there is a one-way road from city i to city j.
Cities are labelled from 1.
For each test case, the first line contains an integer n (1<=n<=500).
The following are n lines, the i-th line contains a string consisting of n characters. If the j-th characters is 1, there is a one-way road from city i to city j.
Cities are labelled from 1.
Output
If there is no solution just output "-1". Otherwise output n integers representing the order to develop this kingdom.
Sample Input
3
011
001
000
0
Sample Output
1 2 3
Source
Mean:
给你一个有n个城市(结点)有向图，现在要发展这些城市，每两个城市之间有且只有一条单向边。
发展城市必须要满足一下条件：
当发展城市w的时候，其他已经发展的城市到w的距离必须小于等于2。
现在要你输出发展的顺序。
analyse:
题目说每两个点之间至少有一条边，所以说数据保证了给的图一定是竞赛图。
由此可知，不能构造的情况是不存在的，也就是说可不能输出-1。
我们只需要对这个图进行一个逆拓扑排序，然后输出就可。
Time complexity：O(n^2)
Source code：
//Memory Time
// 1347K 0MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
char Map[500][500];
int degree[500];
bool vis[500];
int ans[500];
int main ()
{
// freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
int n,i,j,k,l;
while(cin>>n,n)
{
getchar();
for(i=0;i<n;++i)
{
gets(Map[i]);
degree[i]=0;
vis[i]=0;
}
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(Map[i][j]=='1')
degree[j]++;
int vmax,key,idx=-1;
for(int i=0;i<n;i++)
{
vmax=-1,key=-1;
for(int j=0;j<n;j++)
if(!vis[j]&°ree[j]>vmax)
vmax=degree[j],key=j;
vis[key]=1;
ans[++idx]=key;
for(int i=0;i<n;i++)
if(Map[key][i]=='1')
degree[i]--;
}
for(i=n-1;i>=0;--i)
if(i==n-1)
printf("%d",ans[i]+1);
else
printf(" %d",ans[i]+1);
puts("");
}
return 0;
}
没想到这题数据这么水，直接对入度从大到小排序，然后输出就可：
//Memory Time
// 1347K 0MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
char Map[500][500];
struct Node
{
int degree;
int index;
};
Node node[500];
bool cmp(Node a,Node b)
{
return a.degree>b.degree;
}
int main ()
{
// freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
int n,i,j,k,l;
while(cin>>n,n)
{
getchar();
for(i=0;i<n;++i)
{
gets(Map[i]);
node[i].degree=0;
node[i].index=i;
}
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(Map[i][j]=='1')
node[j].degree++;
sort(node,node+n,cmp);
for(i=n-1;i>=0;--i)
if(i==n-1)
printf("%d",node[i].index+1);
else
printf(" %d",node[i].index+1);
puts("");
}
return 0;
}