poj2706 connect

来源：互联网发布：中华人民网络安全法编辑：程序博客网时间：2024/04/29 01:10

唉这题。。线段相交判断+DFS

最终结果的判定就是看有没有赢家，不管谁赢都行；

而且可能给的m步棋没走完，但胜负已分；

只要把m步的能连的边都连起来（先到先连，相同颜色的线也不能相交），然后根据端点是否相等进行DFS，看能不能搜到另一端就好了

偷了几组数据Sample Input4 50 2 2 4 4 2 3 2 2 3 4 50 2 2 4 4 2 3 2 2 1 7 110 3 6 5 3 2 5 7 7 2 4 4 5 3 5 2 4 5 4 0 2 4 5 93 1 3 3 0 2 2 5 2 3 2 1 4 3 4 0 5 1 4 50 1 2 0 1 3 2 2 4 1 8 172 2 2 3 2 4 2 0 3 1 1 6 3 5 4 1 5 1 4 5 5 5 6 0 6 2 6 3 6 4 1 8 4 3 8 212 2 2 3 2 4 2 0 3 1 1 6 3 5 4 1 5 1 4 5 5 5 6 0 6 2 6 3 6 4 1 8 0 1 5 8 8 5 3 7 4 3 8 212 2 2 3 2 4 2 0 3 1 1 6 3 5 4 1 5 1 4 5 5 5 6 0 6 2 6 3 6 4 1 8 0 1 5 8 8 5 4 4 4 3 8 212 2 2 3 2 4 2 0 3 1 1 6 3 5 4 1 5 1 4 5 5 5 6 0 6 2 6 3 6 4 1 8 0 1 5 8 8 3 4 4 4 3 8 212 2 2 3 2 4 2 0 3 1 1 6 3 5 4 1 5 1 4 5 5 5 6 0 6 2 6 3 6 4 1 8 0 1 5 8 8 5 3 3 4 3 10 995 4 7 9 7 3 8 7 1 8 8 10 5 5 5 3 10 7 7 10 4 1 2 2 9 5 3 5 3 9 9 8 4 8 4 2 2 1 2 4 10 4 4 0 0 7 7 0 1 1 9 7 1 7 9 10 0 1 3 7 10 1 7 1 2 3 7 7 7 2 4 5 5 1 6 0 7 4 9 3 3 6 1 4 9 1 2 5 0 9 5 10 4 4 1 3 2 6 1 2 9 9 5 8 4 3 3 10 8 6 1 10 4 7 7 8 6 7 2 10 6 6 5 9 10 6 3 0 5 2 3 3 0 5 8 2 10 5 5 0 4 6 8 8 1 9 2 0 2 9 8 5 0 3 2 8 7 6 1 6 8 1 5 7 7 5 3 4 8 3 6 4 9 6 2 7 0 6 1 0 10 8 6 8 5 6 4 10 10 2 6 1 10 3 4 9 3 1 20 2210 4 8 15 15 3 3 2 1 15 8 11 1 18 6 1 6 4 12 1 16 13 19 1 9 18 5 17 15 12 6 7 19 5 14 15 12 9 6 19 16 6 13 9 3 15 15 13 19 11 9 11 20 14 2 11 12 15 18 4 19 4 12 16 17 10 17 20 2 9 6 20 0 16 19 9 8 18 11 13 14 13 13 17 4 19 14 1 14 12 10 15 6 10 12 10 2 6 17 16 19 18 12 17 13 2 14 2 18 9 17 3 20 5 14 16 7 6 5 5 15 2 6 12 8 6 16 18 1 17 9 16 14 5 8 2 7 1 15 7 15 8 2 1 4 3 1 2 15 4 10 14 10 7 12 19 1 7 3 14 6 15 15 14 11 19 12 4 17 18 6 5 10 2 17 0 9 13 1 5 0 1 13 7 5 16 8 5 9 8 16 3 3 16 18 13 11 6 3 9 8 7 10 19 8 13 15 10 7 4 12 7 8 17 4 14 7 17 11 16 0 13 16 11 6 6 16 2 2 17 8 3 16 15 1 10 2 15 4 2 9 4 12 2 19 8 15 5 14 14 16 9 16 5 9 10 3 7 2 13 7 13 6 9 3 13 10 1 15 18 14 3 2 19 5 0 4 11 7 14 12 3 13 20 17 7 16 17 12 6 18 18 18 14 5 20 12 8 16 0 15 9 5 4 2 3 8 10 20 10 14 7 14 17 8 16 4 5 19 2 10 18 11 18 6 17 11 10 3 8 10 13 13 11 1 6 10 5 7 15 5 12 14 6 7 9 10 3 8 19 8 1 11 4 7 0 13 19 7 12 19 16 17 14 3 12 5 13 12 11 3 0 18 12 3 5 3 19 6 2 0 3 17 17 7 10 13 3 4 15 17 19 20 7 8 8 7 2 10 11 18 17 11 9 15 1 19 0 16 4 13 16 5 8 11 3 20 18 16 19 19 13 18 16 14 19 5 19 15 16 17 2 2 5 9 7 16 7 14 9 0 5 20 2210 4 8 15 15 3 3 2 1 15 8 11 1 18 6 1 6 4 12 1 16 13 19 1 9 18 5 17 15 12 6 7 19 5 14 15 12 9 6 19 16 6 13 9 3 15 15 13 19 11 9 11 20 14 2 11 12 15 18 4 19 4 12 16 17 10 17 20 2 9 6 20 0 16 19 9 8 18 11 13 14 13 13 17 4 19 14 1 14 12 10 15 6 10 12 10 2 6 17 16 19 18 12 17 13 2 14 2 18 9 17 3 20 5 14 16 7 6 5 5 15 2 6 12 8 6 16 18 1 17 9 16 14 5 8 2 7 1 15 7 15 8 2 1 4 3 1 2 15 4 10 14 10 7 12 19 1 7 3 14 6 15 15 14 11 19 12 4 17 18 6 5 10 2 17 0 9 13 1 5 0 1 13 7 5 16 8 5 9 8 16 3 3 16 18 13 11 6 3 9 8 7 10 19 8 13 15 10 7 4 12 7 8 17 4 14 7 17 11 16 0 13 16 11 6 6 16 2 2 17 8 3 16 15 1 10 2 15 4 2 9 4 12 2 19 8 15 5 14 14 16 9 16 5 9 10 3 7 2 13 7 13 6 9 3 13 10 1 15 18 14 3 2 19 5 0 4 11 7 14 12 3 13 20 17 7 16 17 12 6 18 18 18 14 5 20 12 8 16 0 15 9 5 4 2 3 8 10 20 10 14 7 14 17 8 16 4 5 19 2 10 18 11 18 6 17 11 10 3 8 10 13 13 11 1 6 10 5 7 15 5 12 14 6 7 9 10 3 8 19 8 1 11 4 7 0 13 19 7 12 19 16 17 14 3 12 5 13 12 11 3 0 18 12 3 5 3 19 6 2 0 3 17 17 7 10 13 3 4 15 17 19 20 7 8 8 7 2 10 11 18 17 11 9 15 1 19 0 16 4 13 16 5 8 11 3 20 18 16 19 19 13 18 16 14 19 5 19 15 16 17 2 2 5 9 7 16 7 14 9 1 14 10 975 4 7 9 7 3 8 7 1 8 8 10 5 5 5 3 10 7 7 10 4 1 2 2 9 5 3 5 3 9 9 8 4 8 4 2 2 1 2 4 10 4 4 0 0 7 7 0 1 1 9 7 1 7 9 10 0 1 3 7 10 1 7 1 2 3 7 7 7 2 4 5 5 1 6 0 7 4 9 3 3 6 1 4 9 1 2 5 0 9 5 10 4 4 1 3 2 6 1 2 9 9 5 8 4 3 3 10 8 6 1 10 4 7 7 8 6 7 2 10 6 6 5 9 10 6 3 0 5 2 3 3 0 5 8 2 10 5 5 0 4 6 8 8 1 9 2 0 2 9 8 5 0 3 2 8 7 6 1 6 8 1 5 7 7 5 3 4 8 3 6 4 9 6 2 7 0 6 1 0 10 8 6 8 5 6 4 10 10 2 6 1 10 310 995 4 7 9 7 3 8 7 1 8 8 10 5 5 5 3 10 7 7 10 4 1 2 2 9 5 3 5 3 9 9 8 4 8 4 2 2 1 2 4 10 4 4 0 0 7 7 0 1 1 9 7 1 7 9 10 0 1 3 7 10 1 7 1 2 3 7 7 7 2 4 5 5 1 6 0 7 4 9 3 3 6 1 4 9 1 2 5 0 9 5 10 4 4 1 3 2 6 1 2 9 9 5 8 4 3 3 10 8 6 1 10 4 7 7 8 6 7 2 10 6 6 5 9 10 6 3 0 5 2 3 3 0 5 8 2 10 5 5 0 4 6 8 8 1 9 2 0 2 9 8 5 0 3 2 8 7 6 1 6 8 1 5 7 7 5 3 4 8 3 6 4 9 6 2 7 0 6 1 0 10 8 6 8 5 6 4 10 10 2 6 1 10 3 4 9 1 5 0 0 Sample Outputnoyesyesyesnonoyesnoyesnoyesyesnonono

详细看代码。。有注释

#include<iostream>#include<cstdio>#include<cstring>#include<string>#define maxn 1000100#define eps 1e-8using namespace std;typedef struct Point{    int x,y;    friend bool operator == (Point a,Point b)    {        return (a.x==b.x && a.y==b.y);    }}Point;int n,m;struct Edge{    Point u,v;}edge_b[250*250],edge_w[250*250];int edge_num_b,edge_num_w;int mx[]={-2,-2,+2,+2,-1,-1,+1,+1};int my[]={-1,+1,+1,-1,-2,+2,-2,+2};int _map[30][30];bool vis[250*250];bool flag_w,flag_b;double xmult(Point p1,Point p2,Point p){    return (p1.x-p.x)*(p2.y-p.y)-(p2.x-p.x)*(p1.y-p.y);}bool ifcross(Edge A,Edge B){    return  (max(A.u.x,A.v.x) >= min(B.u.x,B.v.x) ) &&            (max(B.u.x,B.v.x) >= min(A.u.x,A.v.x) ) &&            (max(A.u.y,A.v.y) >= min(B.u.y,B.v.y) ) &&            (max(B.u.y,B.v.y) >= min(A.u.y,A.v.y) ) &&            (xmult (A.u,B.u,A.v) * xmult (A.u,A.v,B.v) > eps ) &&            (xmult (B.u,A.u,B.v) * xmult (B.u,B.v,A.v) > eps );}bool cross(int x,int y,int nowx,int nowy){    Edge temp ;    temp.u.x=x;    temp.u.y=y;    temp.v.x=nowx;    temp.v.y=nowy;    int i;    for(i=0;i<edge_num_b;++i)    {        if(ifcross(temp,edge_b[i]))        {            return true;        }    }    for(i=0;i<edge_num_w;++i)    {        if(ifcross(temp,edge_w[i]))            return true;    }    return false;}void add(int x,int y,int type)//1 b 0 w{    int i;    for(i=0;i<8;++i)    {        int nowx=x+mx[i];        int nowy=y+my[i];        if(nowx>=0&&nowx<=n && (type==_map[nowx][nowy]) && !cross(x,y,nowx,nowy) )        {            if(type==1)//black            {                edge_b[edge_num_b].u.x=nowx;                edge_b[edge_num_b].u.y=nowy;                edge_b[edge_num_b].v.x=x;                edge_b[edge_num_b].v.y=y;                edge_num_b++;            }            else            {                edge_w[edge_num_w].u.x=nowx;                edge_w[edge_num_w].u.y=nowy;                edge_w[edge_num_w].v.x=x;                edge_w[edge_num_w].v.y=y;                edge_num_w++;            }        }    }    return;}void dfs_w(Edge temp){    if(flag_w)        return;//    if(temp.v.y==n)//错误    if(temp.v.y==n || temp.u.y==n)//不分方向 能连就行    {        flag_w=1;        return;    }    int i;    for(i=0;i<edge_num_w;++i)    {//        if(!vis[i]&&( (edge_w[i].u==temp.v) ))        if(!vis[i]&&( (edge_w[i].u==temp.v) || (edge_w[i].u==temp.u) || (edge_w[i].v==temp.v) || (edge_w[i].v==temp.u)))//不分方向 能连就行        {            vis[i]=1;            dfs_w(edge_w[i]);//            vis[i]=0;//能走的都走完了 不用撤销标记        }    }}void dfs_b(Edge temp){    if(flag_b)        return;//    if(temp.v.x==n)    if(temp.v.x==n || temp.u.x==n)//不分方向 能连就行    {        flag_b=1;        return;    }    int i;    for(i=0;i<edge_num_b;++i)    {//        if(!vis[i]&& edge_b[i].u==temp.v)        if(!vis[i]&&( (edge_b[i].u==temp.v) || (edge_b[i].u==temp.u) || (edge_b[i].v==temp.v) || (edge_b[i].v==temp.u)))//不分方向 能连就行        {            vis[i]=1;            dfs_b(edge_b[i]);//            vis[i]=0;        }    }}int main(){    while(scanf("%d%d",&n,&m), n||m )    {        int i;        edge_num_w=0;        edge_num_b=0;        memset(_map,-1,sizeof(_map));        int x,y;        for(i=1;i<=m;++i)        {            scanf("%d%d",&x,&y);            if(i&1){//b                _map[x][y]=1;                add(x,y,1);            }            else{                _map[x][y]=0;                add(x,y,0);            }        }        flag_w=0;        memset(vis,false,sizeof(vis));        for(i=0;i<edge_num_w;++i)        {//            if(edge_w[i].u.y==0)            if(edge_w[i].u.y==0||edge_w[i].v.y==0)//不分方向 能连就行            {                vis[i]=1;                dfs_w(edge_w[i]);                vis[i]=0;            }        }        flag_b=0;        memset(vis,false,sizeof(vis));        for(i=0;i<edge_num_b;++i)        {//            if(edge_b[i].u.x==0)            if(edge_b[i].u.x==0||edge_b[i].v.x==0)            {                vis[i]=1;                dfs_b(edge_b[i]);                vis[i]=0;            }        }//        cout<<flag_w<<flag_b<<endl;        if( (flag_b ) || (flag_w ) )        {            printf("yes\n");        }        else        {            printf("no\n");        }    }}

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