Uva 705

题目的意思是在图中寻找可以构成的回路数，及最大回路经过的格子数。

题目不是简单的横竖的格子，而是斜的，这样就不太容易了。后来上网看别人的思路，把图放大三倍，就豁然开朗了。

解题报告链接：http://blog.csdn.net/sio__five/article/details/18910097

代码我修改了一下，留着自己以后慢慢学习。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int length, width, grid, Max;
int g[300][300], isVis[300][300];
int Move[4][2] = {{0, -1},{0, 1} , {-1, 0} ,{1, 0}};

bool IsIn(int x, int y);
void NoEffectFill(int x, int y);
void SetBound();
void Dfs(int x, int y);

int main ()
{
//    freopen("D:\acm.txt","r",stdin);
    int No = 1,cycleNum;
    while(cin>>length>>width,(length||width)){
        width = width * 3, length = length * 3;//将图放大三倍
        memset(g,0,sizeof(g));
        memset(isVis, 0,sizeof(isVis));
        Max = 0;
        cycleNum = 0;
        for (int i = 1; i < width; i += 3){
            for (int j = 1; j < length; j += 3){
                char ch;
                cin>>ch;
                if (ch == '\')
                    g[i - 1][j - 1] = g[i][j] = g[i + 1][j + 1] = 1;
                else
                g[i - 1][j + 1] = g[i][j] = g[i + 1][j - 1] = 1;
            }
        }
        SetBound();
        for (int i = 0; i < width; i++){
            for (int j = 0; j < length; j++){
                if (!g[i][j] && !isVis[i][j]){
                    grid = 0;
                    Dfs(i, j);
                    if (grid >= 12) cycleNum++;//小格子大于12即大格子大于4，构成一个回路
                    if (grid > Max) Max = grid;//记录最大格子数目
                }
            }
        }
        printf("Maze #%d:
", No++);
        if(cycleNum != 0)
            printf("%d Cycles; the longest has length %d.
", cycleNum, Max / 3);
        else
            printf("There are no cycles.
");
        printf("
");
    }
    return 0;
}
bool IsIn(int x, int y)
{
    return (x >= 0 && x < width && y >= 0 && y < length);
}
void NoEffectFill(int x, int y)
{
    int nextX, nextY, i;
    isVis[x][y] = 1, g[x][y] = 2;
    for (i = 0; i < 4; i++){//将外围的空格填充掉
        nextX = x + Move[i][0];
        nextY = y + Move[i][1];
        if (IsIn(nextX, nextY) && !isVis[nextX][nextY] && !g[nextX][nextY])
            NoEffectFill(nextX, nextY);
    }
}
void SetBound()
{
    for (int i = 0; i < width; i++){
        if (!g[i][0] && !isVis[i][0]) 
            NoEffectFill(i, 0);//设置图的外围边界
        if (!g[i][length - 1] && !isVis[i][length - 1]) 
            NoEffectFill(i, length - 1);
    }
    for (int j = 0; j < length; j++){
        if (!g[0][j] && !isVis[0][j])
            NoEffectFill(0, j);//设置图的外围边界
        if (!g[width - 1][j] && !isVis[width - 1][j]) 
            NoEffectFill(width - 1, j);
    }
}
void Dfs(int x, int y)//递归寻找格子
{
    int nextX, nextY;
    isVis[x][y] = 1;
    grid++;
    for (int i = 0; i < 4; i++){
        nextX = x + Move[i][0];
        nextY = y + Move[i][1];
        if (IsIn(nextX, nextY) && !isVis[nextX][nextY] && !g[nextX][nextY])
            Dfs(nextX, nextY);
    }
}