UESTC 889 battle for silver (DFS)

Question:

For a graph, each vertex has a certain authority. Each vertex has at most one edge, and each vertex is connected to at least one vertex through the edge.

We need to find such a group so that all vertices in the group are connected by edges without overlap and the vertices have the largest vertices.

Algorithm:

Because the two connected edges cannot overlap, we can see that there are a maximum of four vertices. So violent search ~

The DFS identifies the elements of the four locations and updates ans while judging whether the points are connected to the previous ones.

At the beginning, I always write at three points and four points. I forgot to consider one point and two points... Wa 10 times ..

If you write a sample yourself, the result is not considered as one or two points ..

/*
6 7
1500
1000
100
2000
500
300
1 2
1 3
1 4
3 5
4 5
4 6
5 6


Ans = 3500
*/

#include<cstdio>#include<iostream>#include<cstring>#include<vector>#define maxn 500using namespace std;int a[maxn],mp[maxn][maxn],v,ans;int op[5];bool vis[maxn];int check(int pos,int x){    for(int i=0;i<pos;i++)    {        if(!mp[op[i]][x])            return 0;    }    return 1;}void dfs(int pos,int x,int sum){    ans = max(ans,sum);    if(pos==4) return;    for(int i=x+1;i<=v;i++)    {        if(vis[i] || !check(pos,i)) continue;        op[pos] = i;        vis[i] = true;        dfs(pos+1,i,sum+a[i]);        vis[i] = false;    }}int main(){    int e,ta,tb;    while(scanf("%d%d",&v,&e)!=EOF)    {        memset(mp,0,sizeof(mp));        for(int i=1;i<=v;i++)            scanf("%d",&a[i]);        for(int i=1;i<=e;i++)        {            scanf("%d%d",&ta,&tb);            mp[ta][tb] = mp[tb][ta] = 1;        }        ans = 0;        dfs(0,0,0);        printf("%d\n",ans);    }    return 0;}

UESTC 889 battle for silver (DFS)