HDU 6029 Graph Theory【水题】
Graph Theory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 448 Accepted Submission(s): 216
Problem Description
Little Q loves playing with different kinds of graphs very much. One day he thought about an interesting category of graphs called ``Cool Graph'', which are generated in the following way:
Let the set of vertices be {1, 2, 3, ..., n}. You have to consider every vertice from left to right (i.e. from vertice 2 to n). At vertice i, you must make one of the following two decisions:
(1) Add edges between this vertex and all the previous vertices (i.e. from vertex 1 to i−1).
(2) Not add any edge between this vertex and any of the previous vertices.
In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices. A perfect matching is a matching that each vertice is covered by an edge in the set.
Now Little Q is interested in checking whether a ''Cool Graph'' has perfect matching. Please write a program to help him.
Let the set of vertices be {1, 2, 3, ..., n}. You have to consider every vertice from left to right (i.e. from vertice 2 to n). At vertice i, you must make one of the following two decisions:
(1) Add edges between this vertex and all the previous vertices (i.e. from vertex 1 to i−1).
(2) Not add any edge between this vertex and any of the previous vertices.
In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices. A perfect matching is a matching that each vertice is covered by an edge in the set.
Now Little Q is interested in checking whether a ''Cool Graph'' has perfect matching. Please write a program to help him.
Input
The first line of the input contains an integer T(1≤T≤50),
denoting the number of test cases.
In each test case, there is an integer n(2≤n≤100000) in the first line, denoting the number of vertices of the graph.
The following line contains n−1 integers a2,a3,...,an(1≤ai≤2), denoting the decision on each vertice.
In each test case, there is an integer n(2≤n≤100000) in the first line, denoting the number of vertices of the graph.
The following line contains n−1 integers a2,a3,...,an(1≤ai≤2), denoting the decision on each vertice.
Output
For each test case, output a string in the first line. If the graph has perfect matching, output ''Yes'', otherwise output ''No''.
Sample Input
3
2
1
2
2
4
1 1 2
Sample Output
Yes
No
No
Source
问题:
有n个点,1表示当前点与之前所有点连一条边,2表示不动,问通过选出其中的某些边,能否使所有的点都能够有一个点与其配对。
思路:
用cnt表示前面有多少个未配对的点,
如果前面有未配对的点则,若操作为1,,则cnt--,若操作为2则cnt++
如果前面所有的点都匹对成功则,若操作为1,,则cnt=1(因为前面没有点与其配对),若操作为2则cnt++
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<ctime>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
#define pi 3.14159265358979
#define mod 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100010;
int n, a[maxn];
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int cnt = 1;
scanf("%d", &n);
for (int i = 2; i <= n; i++)
{
scanf("%d", a + i);
}
if (n % 2 != 0) puts("No");
else
{
for (int i = 2; i <= n; i++)
{
if (a[i] == 1)
{
if (cnt == 0) cnt = 1;
else cnt--;
}
else cnt++;
}
if (cnt > 0) puts("No");
else puts("Yes");
}
}
}
Fighting~