hihoCoder

题意：给你一个三角形的三个坐标，还有一张图，让你从图中的左下角走到右上角，左下角坐标是（0,0），每两点之间的长度是1，'.'代表能走，'#'代表不能走，并且每条路径不能碰到三角形内部，让你求最短路径长度。                                                   

北京错失铜牌，打铁而归，总是感觉很遗憾，也有点难受，这个题当时场上差一丢丢能出的，后来时间没赶上。                    

代码：

 

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <stack>#include <vector>#include <queue>#define ri(n) scanf("%d",&n)#define oi(n) printf("%d\n",n)#define rl(n) scanf("%lld",&n)#define ol(n) printf("%lld\n",n)#define rep(i,l,r) for(i=l;i<=r;i++)#define rep1(i,l,r) for(i=l;i<r;i++)#define eps 1e-8#define zero(x)(((x)>0?(x):-(x))<eps)using namespace std;typedef long long ll;const int inf=0x3f3f3f3f;const int epg=10-8;int vis[1000][1000];int dx[]={-1,0,1,-1,1,-1,0,1};int dy[]={1,1,1,0,0,-1,-1,-1};int n,m;int maps[1000][1000];char s[1000][1000];struct node{    int x,y;    int step;    node(int a,int b,int c):x(a),y(b),step(c){}};struct Point{    double x;    double y;};typedef struct Point point;point p1,p2,p3;double multi(point p0,point p1,point p2)//叉积{    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}bool isIntersected(point s1,point e1,point s2,point e2)//两线段交,不包括端点相交{    return (max(s1.x,e1.x)>=min(s2.x,e2.x)) &&           (max(s2.x,e2.x)>=min(s1.x,e1.x)) &&           (max(s1.y,e1.y)>=min(s2.y,e2.y)) &&           (max(s2.y,e2.y)>=min(s1.y,e1.y)) &&           (multi(s1,s2,e1)*multi(s1,e1,e2)>0) &&           (multi(s2,s1,e2)*multi(s2,e2,e1)>0);}double xmult(point p1,point p2,point p0){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}int dot_online_in(point p,point l1,point l2)//判断点在线段上，包括端点{    return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;}double area(point p1,point p2,point p3)//求三角形面积{    return  abs((p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x))/2.0;}int pansan(point e)//判断点在三角形内部{    if(!dot_online_in(e,p1,p2)&&!dot_online_in(e,p2,p3)&&!dot_online_in(e,p3,p1)&&       (area(e,p1,p2)+area(e,p2,p3)+area(p3,p1,e)<area(p1,p2,p3)+eps))           return 1;    else        return 0;}int same(point p1,point p2){    if(p1.x==p2.x&&p1.y==p2.y)        return 1;    else        return 0;}int pan(point p1,point p2,point p3,point e1,point e2){    if(isIntersected(p1,p2,e1,e2)||isIntersected(p2,p3,e1,e2)||isIntersected(p3,p1,e1,e2))        return 0;    else    {        //int flag1=0,flag2=0;        if((same(e1,p1)&&!same(e2,p2)&&!same(e2,p3)&&dot_online_in(e2,p2,p3))||           (same(e1,p2)&&!same(e2,p1)&&!same(e2,p3)&&dot_online_in(e2,p1,p3))||            (same(e1,p3)&&!same(e2,p1)&&!same(e2,p2)&&dot_online_in(e2,p1,p2))||            (same(e2,p1)&&!same(e1,p2)&&!same(e1,p3)&&dot_online_in(e1,p2,p3))||           (same(e2,p2)&&!same(e1,p1)&&!same(e1,p3)&&dot_online_in(e1,p1,p3))||            (same(e2,p3)&&!same(e1,p1)&&!same(e1,p2)&&dot_online_in(e1,p1,p2))           )            return 0;        else        {            if(((dot_online_in(e1,p1,p2)||dot_online_in(e1,p2,p3)||dot_online_in(e1,p3,p1))&&pansan(e2))               ||((dot_online_in(e2,p1,p2)||dot_online_in(e2,p2,p3)||dot_online_in(e2,p3,p1))&&pansan(e1))               )                return 0;            else            {                if((!same(e1,p1)&&!same(e1,p2)&&dot_online_in(e1,p1,p2)&&( (!same(e2,p1)&&!same(e2,p3)&&dot_online_in(e2,p1,p3))||(!same(e2,p2)&&!same(e2,p3)&&dot_online_in(e2,p2,p3))))||                   (!same(e1,p2)&&!same(e1,p3)&&dot_online_in(e1,p2,p3)&&( (!same(e2,p1)&&!same(e2,p3)&&dot_online_in(e2,p1,p3))||(!same(e2,p2)&&!same(e2,p1)&&dot_online_in(e2,p2,p1))))||                   (!same(e1,p3)&&!same(e1,p1)&&dot_online_in(e1,p3,p1)&&( (!same(e2,p1)&&!same(e2,p2)&&dot_online_in(e2,p1,p2))||(!same(e2,p2)&&!same(e2,p3)&&dot_online_in(e2,p2,p3))))                   )                    return 0;                else                {                     if(pansan(e1)||pansan(e2))                        return 0;                     else                     {                         if(  ((dot_online_in(p1,e1,e2)&&!same(e1,p1)&&dot_online_in(e2,p2,p3)&&!same(e2,p2)&&!same(e2,p3))||(dot_online_in(p1,e1,e2)&&!same(e2,p1)&&dot_online_in(e1,p2,p3)&&!same(e1,p2)&&!same(e1,p3)))                              ||((dot_online_in(p2,e1,e2)&&!same(e1,p2)&&dot_online_in(e2,p1,p3)&&!same(e2,p1)&&!same(e2,p3))||(dot_online_in(p2,e1,e2)&&!same(e2,p2)&&dot_online_in(e1,p1,p3)&&!same(e1,p1)&&!same(e1,p3)))                              ||((dot_online_in(p3,e1,e2)&&!same(e1,p3)&&dot_online_in(e2,p1,p2)&&!same(e2,p1)&&!same(e2,p2))||(dot_online_in(p3,e1,e2)&&!same(e2,p3)&&dot_online_in(e1,p1,p2)&&!same(e1,p1)&&!same(e1,p2)))                            )                                return 0;                         else                            return 1;                     }                        //return 1;                }            }                //return 1;        }    }}int bfs(){memset(vis,false,sizeof(vis));queue<node> q;vis[0][0] = true;q.push(node(0,0,0));while(!q.empty()){node p = q.front();q.pop();int nx,ny;for(int i = 0 ; i < 8 ; i ++){nx = p.x + dx[i];ny = p.y + dy[i];if(nx < 0 || nx >= n || ny < 0 || ny >= n)continue;if(maps[nx][ny] == '#' || vis[nx][ny])continue;Point q1,q2;q1.x = p.x;q1.y = p.y;q2.x = nx;q2.y = ny;//cout << "p.x = " << p.x << " p.y = " << p.y << endl;if(pan(p1,p2,p3,q1,q2)){vis[nx][ny] = true;//cout << "nx = " << nx << " ny = " << ny << endl;q.push(node(nx,ny,p.step+1));if(nx == n-1 && ny == n-1){return p.step+1;}}}}return -1;}int main(){    while(scanf("%d",&n) != EOF){    cin>>p1.x>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y;    //cin>>p1.y>>p1.x>>p2.y>>p2.x>>p3.y>>p3.x;    for(int i = 0 ; i < n ; i ++){    scanf("%s",s[i]);    }    for(int i = 0 ; i < n ; i ++){    for(int j = 0 ; j < n ; j ++){    maps[i][j] = s[n-1-j][i];    //maps[i][j] = s[i][j];    }    }    // for(int i = 0 ; i < n ; i ++){    // for(int j = 0 ; j < n ; j ++){    // cout << maps[i][j];    // }    // cout << endl;    // }    cout << bfs() << endl;    }    return 0;}