hdu 1045(dfs暴搜/缩点+二分图最大匹配（典型行列匹配题）)

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题目链接：Fire Net

题目大意：在一个方格图上放最多的大炮，指的每两个大炮（横竖）至少有一份blockhose将他们隔开

题目分析：

一、dfs暴搜

参见八皇后问题，（搜索其四个方向）判断每一个格子能不能放，能放，就标记为'@'

因为棋盘大小最多为4，所以不会TLE

#include <stdio.h>#include <memory.h>using namespace std;char g[5][5];int tot, sz, ans;int dx[] = { -1, 1, 0, 0 };int dy[] = { 0, 0, -1, 1 };bool check(int x, int y){    if (x<0 || x>sz - 1 || y<0 || y>sz - 1)return false;    else return true;}bool judge(int x, int y){    bool flag = true;    for (int i = 0; i<4; ++i){        if (!flag)return flag;        int nx = x + dx[i];        int ny = y + dy[i];        while (check(nx, ny)){            if (g[nx][ny] == '@'){                flag = false;                break;            }            else if (g[nx][ny] == 'X')break;            nx = nx + dx[i];            ny = ny + dy[i];        }    }    return flag;}void dfs(int cur){    if (cur > ans)ans = cur;    for (int i = 0; i < sz; ++i){        for (int j = 0; j < sz; ++j){            if (g[i][j] == '.'){                if (g[i][j] == '.'&&judge(i, j)) {                    g[i][j] = '@';                    dfs(cur + 1);//放                    g[i][j] = '.';                }            }        }    }}int main(){    //freopen("in.txt", "r", stdin);    while (scanf("%d", &sz) != EOF&&sz) {        ans = 0;        for (int i = 0; i < sz; ++i){            scanf("%s", g[i]);        }        dfs(0);        printf("%d\n", ans);    }    return 0;}

二、二分图最大匹配

因为没有被墙隔开的只能放一个，因此按图所示，连续的空格可以缩成一个点

横着的点和竖着的点重合的地方都是可以放的，

相当于将横点和竖点匹配起来->二分图匹配

代码

#include <stdio.h>#include <memory.h>int sz;char str[5][5];int gRow[5][5], gCol[5][5],g[8][8];int nRow, nCol;bool vis[8];int link[8];int dfs(int u){    for (int v = 0; v < nCol; ++v){        if (g[u][v] == 1 && !vis[v]){            vis[v] = true;            if (link[v] == -1 || dfs(link[v])){                link[v] = u;                return 1;            }        }    }    return 0;}int hungry(){    int res = 0;    memset(link, -1, sizeof(link));    for (int i = 0; i < nRow; ++i){        memset(vis, 0, sizeof(vis));        res += dfs(i);    }    return res;}int main(){    //freopen("in.txt", "r", stdin);    while (scanf("%d", &sz) != EOF&&sz){        memset(gRow, -1, sizeof(gRow));        memset(gCol, -1, sizeof(gCol));        memset(g, -1, sizeof(g));        for (int i = 0; i < sz; ++i){            scanf("%s", str[i]);        }        nRow = 0, nCol = 0;        for (int i = 0; i < sz; ++i){            for (int j = 0; j < sz; ++j){                if (str[i][j] == '.'){                    int k;                    for (k = j; str[i][k] == '.'&&k < sz; ++k) {                        gRow[i][k] = nRow;                    }                    j = k;                    nRow++;                }            }        }        for (int j = 0; j < sz; ++j){            for (int i = 0; i < sz; ++i){                if (str[i][j] == '.'){                    int k;                    for (k = i; str[k][j] == '.'&&k < sz; ++k){                        gCol[k][j] = nCol;                    }                    i = k;                    nCol++;                }            }        }        for (int i = 0; i < sz; ++i){            for (int j = 0; j < sz; ++j){                if(str[i][j]=='.')g[gRow[i][j]][gCol[i][j]] = 1;            }        }        printf("%d\n", hungry());    }    return 0;}

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