uva 229 - Scanner（枚举技巧）

题目链接：uva 229 - Scanner

题目大意：工厂有一个扫描器，从四条线进行扫描，它可以扫描出直线上有多少个位置是存在物体的。请还原出物体的形状，如果不能确定或者矛盾，输出空点。

解题思路：由确定的某条线去修改没有确定的线，然后会产生出新的确定线，以此计算，如果最后确定线的个数不等于总的个数，则是矛盾。

#include <stdio.h>
#include <string.h>

const int cN[4] = {10, 24, 15, 24};
const int N = 30;

int S[4]
, tB[4]
, tW[4]
;
int cB[4]
, cW[4]
, g

, v[4]
;

void init () {
for (int i = 0; i < cN[0]; i++)
S[0][i] = 15;

for (int i = 0; i < cN[1]; i++) {
if (i < 10)
S[1][i] = i + 1;
else if (i > 13)
S[1][i] = 24 - i;
else
S[1][i] = 10;
}

for (int i = 0; i < cN[2]; i++)
S[2][i] = 10;

for (int i = 0; i < cN[3]; i++) {
if (i < 10)
S[3][i] = i + 1;
else if (i > 13)
S[3][i] = 24 - i;
else
S[3][i] = 10;
}
}

void read () {
memset(g, -1, sizeof(g));
memset(v, 0, sizeof(v));
memset(cB, 0, sizeof(cB));
memset(cW, 0, sizeof(cW));

for (int i = 0; i < 4; i++) {
for (int j = 0; j < cN[i]; j++) {
scanf("%d", &tB[i][j]);
tW[i][j] = S[i][j] - tB[i][j];
}
}
}

void cat (int x, int y) {
if (g[x][y] == 1) {
cB[0][x]++; cB[1][x+y]++; cB[2][y]++; cB[3][y-x+9]++;
} else {
cW[0][x]++; cW[1][x+y]++; cW[2][y]++; cW[3][y-x+9]++;
}
}

void draw (int x, int y, int val) {
v[x][y] = 1;

if (x == 0) {

for (int i = 0; i < 15; i++) {
if (g[y][i] == -1) {
g[y][i] = val;
cat (y, i);
}
}
} else if (x == 1) {

for (int i = 0; i < 10; i++) {
for (int j = 0; j < 15; j++) {
if (i + j != y) continue;
if (g[i][j] == -1) {
g[i][j] = val;
cat (i, j);
}
}
}
} else if (x == 2) {

for (int i = 0; i < 10; i++) {
if (g[i][y] == -1) {
g[i][y] = val;
cat (i, y);
}
}
} else {

for (int i = 0; i < 10; i++) {
for (int j = 0; j < 15; j++) {
if (j - i  + 9 != y) continue;
if (g[i][j] == -1) {
g[i][j] = val;
cat (i, j);
}
}
}
}
}

bool judge () {
for (int i = 0; i < 4; i++) {
for (int j = 0; j < cN[i]; j++) {
if (v[i][j]) continue;

if (cW[i][j] == tW[i][j]) {
draw (i, j, 1);
return true;
} else if (cB[i][j] == tB[i][j]) {
draw (i, j, 0);
return true;
}
}
}
return false;
}

void solve () {
int cnt = 0;
while (judge())
cnt++;

if (cnt != 73) memset(g, 0, sizeof(g));

for (int i = 0; i < 10; i++) {
for (int j = 0; j < 15; j++)
printf("%c", g[i][j] ? '#' : '.');
printf("\n");
}
}

int main () {
init ();

int cas;
scanf("%d", &cas);

while (cas--) {
read ();
solve ();
if (cas) printf("\n");
}
return 0;
}