Usaco Section 2.1 The Castle

最新推荐文章于 2024-05-04 16:56:52 发布

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最新推荐文章于 2024-05-04 16:56:52 发布

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分类专栏： Algorithm Competition 文章标签： null struct 算法 file c

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本文通过解决USACO的城堡问题，介绍了Floodfill算法的应用。该算法用于分割地图上的不同房间，并计算每个房间的大小及最大可能的空间合并方案。

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题目：http://ace.delos.com/usacoprob2?a=eURSI1UXlof&S=castle

floodfill算法的典型应用！

ID: lvxiaol3

PROG: castle

LANG: C

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <assert.h>

#define MAXDIM 50

#define MAXN 100

#define MAXCOLOR 100

#define MAXROOM (MAXDIM*MAXDIM)

enum

{

Wwest = 1,

Wnorth = 2,

Weast = 4,

Wsouth = 8

};

typedef struct Square Square;

struct Square

{

int wall;

int numbered; //是否被遍历过

int room;

};

int wid, ht;

Square castle[MAXDIM][MAXDIM];

int roomsize[MAXROOM];

void number(int room, int x, int y)

{

int w;

if(castle[x][y].numbered)

{

assert(castle[x][y].room == room);

return;

}

castle[x][y].numbered = 1;

castle[x][y].room = room;

roomsize[room]++;

w = castle[x][y].wall;

if(x > 0 && !(w & Wwest)) //只有x>0&&向西有空房间 number(room,x-1,y)

number(room, x-1, y);

if(x+1 < wid && !(w & Weast))

number(room, x+1, y);

if(y > 0 && !(w & Wnorth))

number(room, x, y-1);

if(y+1 < ht && !(w & Wsouth))

number(room, x, y+1);

}

void main(void)

{

FILE *fin, *fout;

int x, y, w, nroom, bigroom;

int i, n, m, mx, my;

char mc;

fin = fopen("castle.in", "r");

fout = fopen("castle.out", "w");

assert(fin != NULL && fout != NULL);

fscanf(fin, "%d %d", &wid, &ht);

/* read in wall info */

for(y=0; y<ht; y++)

{

for(x=0; x<wid; x++)

{

fscanf(fin, "%d", &w);

castle[x][y].wall = w;

}

/* number rooms */

nroom = 0;

for(y=0; y<ht; y++)

for(x=0; x<wid; x++)

if(!castle[x][y].numbered)

number(nroom++, x, y);

/* find biggest room */

bigroom = roomsize[0];

for(i=1; i<nroom; i++)

if(bigroom < roomsize[i])

bigroom = roomsize[i];

/* look at best that can come of removing an east or north wall */

//下面的策略是遍历了整个空间，来寻找最大的，并用m保存。

m = 0;

for(x=0; x<wid; x++)

{

for(y=ht-1; y>=0; y--)

{

if(y > 0 && castle[x][y].room != castle[x][y-1].room)

{

n = roomsize[castle[x][y].room] + roomsize[castle[x][y-1].room];

if(n > m)

{

m = n;

mx = x;

my = y;

mc = 'N';

}

if(x+1 < wid && castle[x][y].room != castle[x+1][y].room)

{

n = roomsize[castle[x][y].room] + roomsize[castle[x+1][y].room];

if(n > m)

{

m = n;

mx = x;

my = y;

mc = 'E';

}

fprintf(fout, "%d/n", nroom);

fprintf(fout, "%d/n", bigroom);

fprintf(fout, "%d/n", m);

fprintf(fout, "%d %d %c/n", my+1, mx+1, mc);

exit(0);

}

//这个问题精彩的地方在于

/*nroom = 0;

for(y=0; y<ht; y++)

for(x=0; x<wid; x++)

if(!castle[x][y].numbered)

number(nroom++, x, y);

//这一段的使用把整体的房间分割并标号。