SPOJ IM 962 网络流最大流 解题报告

M - Intergalactic Map

Map Jedi knights, Qui-Gon Jinn and his young apprentice Obi-Wan Kenobi, are entrusted by Queen Padmé Amidala to save Naboo from an invasion by the Trade Federation. They must leave Naboo immediately and go to Tatooine to pick up the proof of the Federation’s evil design. They then must proceed on to the Republic’s capital planet Coruscant to produce it in front of the Republic’s Senate. To help them in this endeavor, the queen’s captain provides them with an intergalactic map. This map shows connections between planets not yet blockaded by the Trade Federation. Any pair of planets has at most one connection between them, and all the connections are two-way. To avoid detection by enemy spies, the knights must embark on this adventure without visiting any planet more than once. Can you help them by determining if such a path exists?

Note - In the attached map, the desired path is shown in bold.

Input Description

The first line of the input is a positive integer t ≤ 20, which is the number of test cases. The descriptions of the test cases follow one after the other. The first line of each test case is a pair of positive integers n, m (separated by a single space). 2 ≤ n ≤ 30011 is the number of planets and m ≤ 50011 is the number of connections between planets. The planets are indexed with integers from 1 to n. The indices of Naboo, Tatooine and Coruscant are 1, 2, 3 respectively. The next m lines contain two integers each, giving pairs of planets that have a connection between them.

Output Description

The output should contain t lines. The ith line corresponds to the ith test case. The output for each test case should be YES if the required path exists and NO otherwise.

Example

Input

2

3 3

1 2

2 3

1 3

3 1

1 3

Output

YES

NO

【解题报告】

【题目大意】

在一个无向图中，一个人要从A点赶往B点，之后再赶往C点，且要求中途不能多次经过同一个点。问是否存在这样的路线。（3 <= N <= 30011, 1 <= M <= 50011）

【建模方法】

由于每个点只能走一次，似乎最短路之类的算法不能用，只有往网络流上靠。将每个点i拆成两个点i’, i’’并加边(i’, i’’, 1)就能轻易达到这个目的。起初我一直以A为源点思考，却怎么也想不出如何处理先后经过两个汇点的问题，直到灵光一现，想到可以以B为源点，A、C为汇点。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define N 30020
#define M 50020
#define inf 0x3f3f3f3f

int cnt,head[N],dis[N],vis[N];
struct Edge{int to,nxt,f;}e[M<<];
int t,n,m,ss,tt;

void adde(int u,int v,int c)
{
    e[++cnt].to=v;e[cnt].f=c;
    e[cnt].nxt=head[u];head[u]=cnt;
    e[++cnt].to=u;e[cnt].f=;
    e[cnt].nxt=head[v];head[v]=cnt;
}
bool bfs()
{
    memset(dis,,sizeof(dis));
    memset(vis,,sizeof(vis));
    queue <int> q;
    vis[ss]=;q.push(ss);
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];~i;i=e[i].nxt)
        {
            int v=e[i].to;
            if(e[i].f&&!vis[v])
            {
                q.push(v);vis[v]=;
                dis[v]=dis[u]+;
            }
        }
    }
    return vis[tt];
}
int dfs(int u,int delta)
{
    if(u==tt) return delta;
    int ret=;
    for(int i=head[u];delta&&~i;i=e[i].nxt)
    {
        int v=e[i].to;
        if(e[i].f&&dis[v]==dis[u]+)
        {
            int flow=dfs(v,min(e[i].f,delta));
            e[i].f-=flow;
            e[i^].f+=flow;
            delta-=flow;
            ret+=flow;
        }
    }
    return ret;
}
int Dinic()
{
    int ret=;
    while(bfs()) ret+=dfs(ss,inf);
    return ret;
}
int main()
{
    for(scanf("%d",&t);t;--t)
    {
        cnt=-;
        memset(head,-,sizeof(head));
        scanf("%d%d",&n,&m);
        ss=,tt=n*+;
        adde(ss,+n,);
        adde(,tt,);
        adde(,tt,);
        for(int i=;i<=n;++i) adde(i,i+n,);
        for(int i=,u,v;i<=m;++i)  
        {  
            scanf("%d%d",&u,&v);  
            if(u<||u>n||v<||v>n) continue;  
            adde(u+n,v,);adde(v+n,u,);  
        }     
        puts(Dinic()==?"YES":"NO");
    }
    return ;
}