Rooks

Rooks

时间： 1ms        内存：128M

描述：

You have unexpectedly become the owner of a large chessboard, having fifteen squares to each side. Because you do not know how to play chess on such a large board, you find an alternative way to make use of it.

In chess, a rook attacks all squares that are in the same row or column of the chessboard as it is. For the purposes of this problem, we define a rook as also attacking the square on which it is already standing.

Given a set of chessboard squares, how many rooks are needed to attack all of them?

输入：

Input consists of a number of test cases. Each test case consists of fifteen lines each containing fifteen characters depicting the chess board. Each character is either a period (.) or a hash (#). Every chessboard square depicted by a hash must be attacked by a rook. After all the test cases, one more line of input appears. This line contains the word END.

输出：

Output consists of exactly one line for each test case. The line contains a single integer, the minimum number of rooks that must be placed on the chess board so that every square marked with a hash is attacked.

示例输入：

...............
...............
...............
...............
...............
...............
...............
.......#.......
...............
...............
...............
...............
...............
...............
...............
END

示例输出：

1

提示：

参考答案（内存最优[800]）：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define FR(i,a,b) for(int i=(a);i<(b);++i)
#define FOR(i,n) FR(i,0,n)
#define CLR(x,a) memset(x,a,sizeof(x))
#define setmax(a,b) a = max(a,b)
#define setmin(a,b) a = min(a,b)
typedef long long ll;

const int R=15,C=15;
int i;
char grid[16][16];
int colset[16];
void doit() {
  FOR(r,R) {
    scanf("%s",grid[r]);
    if (0==strcmp("END",grid[r])) exit(0);
  }

  int ans = 15;
  FOR(S,(1<<R)) {
    CLR(colset,0);

    FOR(r,R) if (0==((1<<r)&S)) {
      FOR(c,C) if (grid[r][c]=='#') colset[c]=1;
    }

    int now = 0;
    FOR(c,C) now += colset[c];
    setmax(now, __builtin_popcount(S));
    setmin(ans, now);
  }

  printf("%d\n",ans);
}
int main() {
  while (1) doit();
}

参考答案（时间最优[20]）：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <string>
#include <vector>
using namespace std;

#define SZ(v) int((v).size())
#define FR(i,a,b) for(int i=(a);i<(b);++i)
#define FOR(i,n) FR(i,0,n)
#define FORI(i,v) FOR(i,SZ(v))
#define CLR(x,a) memset(x,a,sizeof(x))
#define setmax(a,b) a = max(a,b)
#define setmin(a,b) a = min(a,b)
typedef long long ll;

int R,C;
char buf[16];
int deg[16][16];
int nod[16][16][4];
const int MAXV = 514;
bool mark[MAXV];
int cap[MAXV][MAXV];
vector<int> edg[MAXV];
const int s=MAXV-2,t=MAXV-1;
void connect(int v, int w, int u) {
  cap[v][w] = u;
  cap[w][v] = 0;
  edg[v].push_back(w);
  edg[w].push_back(v);
}
int dr[] = { -1, 0, 1, 0 },
    dc[] = { 0, 1, 0, -1 };
const int inf = 123456789;
int dfs(int v, int flocap=inf) {
  if (v==t) return flocap;

  if (mark[v]) return 0;
  mark[v] = 1;

  FORI(i,edg[v]) {
    int w = edg[v][i];
    if (cap[v][w]) {
      int f = dfs(w, min(flocap, cap[v][w]));

      if (f) {
	cap[v][w] -= f;
	cap[w][v] += f;
	return f;
      }
    }
  }
  return 0;
}
void doit() {
  scanf("%d%d",&R,&C);
  if (R==0) exit(0);

  FOR(v,MAXV) edg[v].clear();
  CLR(cap,0);

  FOR(r,R) {
    FOR(c,C) {
      scanf(" %s",buf);
      if (0==strcmp("x",buf)) deg[r][c]=0;
      else deg[r][c] = strlen(buf);
    }
  }

  int next = 0;
  FOR(r,R) FOR(c,C) {
    if (deg[r][c] != 2) {
      FOR(k,4) nod[r][c][k] = next;
      if ((r+c)%2) connect(s, next, deg[r][c]);
      else connect(next, t, deg[r][c]);
      ++next;
    } else {
      FOR(k,4) nod[r][c][k] = next + (k%2);
      if ((r+c)%2) FOR(k,2) connect(s, next+k, 1);
      else FOR(k,2) connect(next+k, t, 1);
      next += 2;
    }
  }

  FOR(r,R) FOR(c,C) if ((r+c)%2) {
    FOR(k,4) {
      int r2 = r+dr[k], c2 = c+dc[k];
      if (r2<0||r2>=R||c2<0||c2>=C) continue;
      int k2 = (k+2)%4;

      connect(nod[r][c][k], nod[r2][c2][k2], 1);
    }
  }

  while (1) {
    CLR(mark,0);
    int f = dfs(s);
    if (f == 0) break;
  }

  int ed=0,od=0;
  FOR(r,R) FOR(c,C) {
    if ((r+c)%2) od += deg[r][c];
    else ed += deg[r][c];
  }

  bool happy = 1;
  FOR(v,MAXV) if (cap[s][v] || cap[v][t]) happy = 0;

  printf("%s\n", happy ? "SOLVABLE" : "UNSOLVABLE");
}
int main() {
  while (1) doit();
}

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