（HDU 6029 女生专场）Graph Theory 思维题

Graph Theory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 346 Accepted Submission(s): 167

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.

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.

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
2017中国大学生程序设计竞赛 - 女生专场

题意：
给你一个有n个节点的图，编号为1,2,..,n。图的边的信息有两种情况：
若节点i的信息是1：表示i和1~i-1的所有的节点都有一条边
若节点i的信息是2：表示i和1~i-1的所有的节点都没有
一个匹配：在一个图的边的集合中没有相同的节点
一个完美匹配：所有的节点都在这个集合中
现在问你，这个图是否是完美匹配的

分析：
由于一个节点，要么和前的节点都有边，要么都没有，所以对于所有后面为1的节点，表示它可以和前面的任意节点进行匹配，而后面为2的节点，要想完美匹配它一定会和它后面为1的节点进行匹配。
相当于后面为1的节点，给你所有节点一个匹配的机会，而后面为2的节点，要占用一次机会。
所有我们用sum记录机会的次数，当遇到1时sum++遇到2时sum–.只要满足在这个过程中sum>=0 ,并且最后sum >=1 (第一个节点为2)即可。

注意：n为奇数时一定不行

由于我判断n为奇数后就直接出了结果，没有进行读取数据了导致wa了几次。。。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;const int maxn = 100010;int a[maxn];int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<n;i++)        {            scanf("%d",&a[i]);            if(a[i] == 2) a[i] = -1;        }        if((n%2) == 1)        {            printf("No\n");            continue;        }        int sum = 0;        int f = 1;        for(int i=n-1;i>=1;i--)        {            sum += a[i];            if(sum < 0)            {                f = 0;                break;            }        }        if(f == 0 || sum < 1) printf("No\n");        else printf("Yes\n");    }    return 0;}