USACO 1.3 Wormholes

这题卡了我n久时间。。。

Solution:

1. 按照坐标y值排序，同时按x从小到大排序
2. 连边: 纵坐标相同的点从左至右连边；eg: dot1（1,3）dot2（3,3）dot3（5,3）连dot1->dot2,dot2->3
3. DFS选择不同的连边组合，边数达到n/2时判断并结束
4. 输出结果

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/*ID: zzzc181PROG: wormholeLANG: C++*/#include<ctime>#include<cstdio>#include<cstring>#include<algorithm>#define MAXN 30using namespace std;struct data{    int x,y,k;}node[MAXN];int n,ans;int vis[MAXN];int edge1[MAXN],edge2[MAXN];bool cmp(const data & a, const data & b){    if(a.y!=b.y){        return a.y<b.y;    }    return a.x<b.x;}bool check(int x){    memset(vis,0,sizeof(vis));    bool pd=false;    while(1){        vis[x]++;        if(vis[x]>n) return true;        int i;        pd=false;        if(edge1[x]){            x=edge2[edge1[x]];            pd=true;        }        if(!pd) return false;    }    return false;}void DFS(int x){    if(x==n/2+1){        int p;        for(p=1;p<=n;p++){            if(check(p)){                ans++;                return;            }        }        return;    }    int i,j;    for(i=1;i<=n;i++){        if(!edge2[i]){            for(j=i+1;j<=n;j++){                if(!edge2[j]){                    edge2[i]=j;edge2[j]=i;                    DFS(x+1);                    edge2[i]=0;edge2[j]=0;                }            }            break;        }    }}void solve(){    int i;    sort(node+1,node+n+1,cmp);    for(i=2;i<=n;i++){        if(node[i-1].y==node[i].y){            edge1[i-1]=i;        }    }    DFS(1);    printf("%d\n",ans);}int main(){//  clock_t t1=clock();    freopen("wormhole.in","r",stdin);    freopen("wormhole.out","w",stdout);    scanf("%d",&n);    int i;    for(i=1;i<=n;i++){        scanf("%d%d",&node[i].x,&node[i].y);        node[i].k=i;    }    solve();//  clock_t t2=clock();//  printf("%lf\n",(double)(t2-t1)/CLOCKS_PER_SEC);    return 0;}