poj 1970(搜索)

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Description

A game of Renju is played on a 19*19 board by two players. One player uses black stones and the other uses white stones. The game begins in an empty board and two players alternate in placing black stones and white stones. Black always goes first. There are 19 horizontal lines and 19 vertical lines in the board and the stones are placed on the intersections of the lines.

Horizontal lines are marked 1, 2, ..., 19 from up to down and vertical lines are marked 1, 2, ..., 19 from left to right.

The objective of this game is to put five stones of the same color consecutively along a horizontal, vertical, or diagonal line. So, black wins in the above figure. But, a player does not win the game if more than five stones of the same color were put consecutively.

Given a configuration of the game, write a program to determine whether white has won or black has won or nobody has won yet. There will be no input data where the black and the white both win at the same time. Also there will be no input data where the white or the black wins in more than one place.

Input

Output

Sample Input

1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 2 0 0 2 2 2 1 0 0 0 0 0 0 0 0 0 0
0 0 1 2 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 2 2 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sample Output

1
3 2

错了N次。。这个题坑点在于只能是五子棋，6子，7子都不行，所以对一个点的某一个方向来说正反都要搜一遍。
而且还要注意是结果是要位于左上角的点。所以可以先将某一列的每一行先找一遍，这样的话得到的结果就一定是左上角的点。
给大家两组测试用例：
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 2 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
输出是0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 2 2 2 1 0 0 0 0 0 0 0 0 0 0
0 0 1 2 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 1 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 1 2 2 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
答案是
1
6 1

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
int graph[20][20];
bool vis[20][20];
int cnt;
int res,resx,resy;
int dir[][2] = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,-1},{-1,1},{1,-1}};
bool check(int x,int y,int flag)
{
    if(x<1||x>19||y<1||y>19||graph[x][y]!=flag) return false;
    return true;
}
struct Node
{
    int x,y;
    int step;
};
Node s;
bool bfs(int x,int y,int flag)
{
    Node now;
    now.x = x,now.y = y,now.step = 0;
    Node next;
    for(int i=0; i<8; i++)
    {
        next.x = now.x+dir[i][0];
        next.y = now.y +dir[i][1];
        next.step = now.step+1;
        while(check(next.x,next.y,flag))
        {
            next.x+=dir[i][0];
            next.y+=dir[i][1];
            next.step++;
        }
        int step1 = next.step - 1;
        next.x = now.x+dir[i^1][0]; ///反方向也要找
        next.y = now.y +dir[i^1][1];
        next.step = now.step+1;
        while(check(next.x,next.y,flag))
        {
            next.x+=dir[i^1][0];
            next.y+=dir[i^1][1];
            next.step++;
        }
        int step2 = next.step - 1;
        if(step1+step2==4) return true;
    }
    return false;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        for(int i=1; i<20; i++)
        {
            for(int j=1; j<20; j++)
            {
                scanf("%d",&graph[i][j]);
            }
        }
        bool flag = false;
        res = 0,resx=-1,resy=-1;
        for(int j=1; j<20&&!flag; j++)
        {
            for(int i=1; i<20&&!flag; i++)
            {
                if(graph[i][j]==1)
                {
                    flag =  bfs(i,j,1);
                    if(flag)
                    {
                        res = 1;
                        resx = i;
                        resy = j;
                    }
                }
                if(graph[i][j]==2)
                {
                    flag =  bfs(i,j,2);
                    if(flag)
                    {
                        res = 2;
                        resx = i;
                        resy = j;
                    }
                }
            }
        }
        if(res==0) printf("0\n");
        else printf("%d\n%d %d\n",res,resx,resy);
    }
    return 0;
}

poj 1970(搜索)

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原文地址：http://www.cnblogs.com/liyinggang/p/5574518.html