#include<iostream>
#include<vector>
#include<string>
using namespace std;
void chushihua
(int r,
int c,vector<string> &result,vector< vector<
int> > &newresult,vector<vector<
int> > &noplace
){
for
(int i=0;i<r*c;i
++){
if
(result[0][i]=='X'
){
newresult[i/c][i%c]=7;
noplace[i/c][i%c]=1;
}
else if
(result[0][i]=='0'
){
newresult[i/c][i%c]=0;
noplace[i/c][i%c]=1;
}
else if
(result[0][i]=='1'
){
newresult[i/c][i%c]=1;
noplace[i/c][i%c]=1;
}
else if
(result[0][i]=='2'
){
newresult[i/c][i%c]=2;
noplace[i/c][i%c]=1;
}
else if
(result[0][i]=='3'
){
newresult[i/c][i%c]=3;
noplace[i/c][i%c]=1;
}
else if
(result[0][i]=='4'
){
newresult[i/c][i%c]=4;
noplace[i/c][i%c]=1;
}
else{
newresult[i/c][i%c]=6;
noplace[i/c][i%c]=0;
}
}
}
// 放置灯的函数（4方向，光线遇到墙壁 7 或数字 0~4 停止）
void placeLight
(int r,
int c, vector<vector<
int> >& newresult, vector<vector<
int> >& noplace
) {
int rows = newresult.size
();
if
(rows == 0
) return;
int cols = newresult[0].size
();
// 检查是否可以放灯：必须是空地 6，且当前允许放置（noplace == 0）
if
(r < 0 || r >= rows || c < 0 || c >= cols
) return;
if
(newresult[r][c] != 6 || noplace[r][c] != 0
) return;
// 放置灯
newresult[r][c] = 5; // 5 代表灯
// 四个方向：上
(-1,0
)、下
(1,0
)、左
(0,-1
)、右
(0,1
)
int dr[] = {-1, 1, 0, 0};
int dc[] = {0, 0, -1, 1};
for
(int d = 0; d < 4;
++d
) {
int dir_r = dr[d];
int dir_c = dc[d];
int nr = r
+ dir_r;
int nc = c
+ dir_c;
while
(true
) {
//
越界，停止
if
(nr < 0 || nr >= rows || nc < 0 || nc >= cols
) {
break;
}
int cell = newresult[nr][nc];
// 遇到墙壁 7 或数字 0~4，光线停止
if
(cell == 7 ||
(cell >= 0 && cell <= 4
)) {
break;
}
// 如果是可放置的空格 6，则在 noplace 中标记为不能放灯
if
(cell == 6
) {
noplace[nr][nc] = 1;
}
// 沿当前方向继续前进
nr
+= dir_r;
nc
+= dir_c;
}
}
}
// 获取
(r,c
) 的上下左右 4 个邻居，存入 neighbors（每个元素是 pair<
int,
int>）
void getNeighbors
(int r,
int c,
int rows,
int cols, vector<pair<
int,
int> >& neighbors
) {
int dr[] = {-1, 1, 0, 0}; // 上，下
int dc[] = {0, 0, -1, 1}; // 左，右
for
(int d = 0; d < 4; d
++) {
int nr = r
+ dr[d];
int nc = c
+ dc[d];
if
(nr >= 0 && nr < rows && nc >= 0 && nc < cols
) {
neighbors.push_back
(make_pair
(nr, nc
));
}
}
}
void canreason
(int rows,
int cols, vector<vector<
int> >& newresult, vector<vector<
int> >& noplace
) {
// ===== 规则 1：如果数字是 4，就往上下左右放灯（如果可放）=====
for
(int i = 0; i < rows; i
++) {
for
(int j = 0; j < cols; j
++) {
if
(newresult[i][j] == 4
) {
vector<pair<
int,
int> > neighbors;
getNeighbors
(i, j, rows, cols, neighbors
);
vector<pair<
int,
int> > available_spots;
// 遍历 4 个邻居
for
(int k = 0; k < neighbors.size
(); k
++) {
pair<
int,
int> p = neighbors[k];
int nr = p.first;
int nc = p.second;
int val = newresult[nr][nc];
if
(val == 6 && noplace[nr][nc] == 0 && val != 5 && val != 7
) {
available_spots.push_back
(make_pair
(nr, nc
));
}
}
// 如果刚好有 4 个可放的空位，就都放灯
if
(available_spots.size
() == 4
) {
for
(int k = 0; k < available_spots.size
(); k
++) {
pair<
int,
int> p = available_spots[k];
int nr = p.first;
int nc = p.second;
placeLight
(nr, nc, newresult, noplace
);
}
}
}
}
}
// ===== 规则 2：如果数字是 0~3，且已放的灯数 == 数字，则放周围可放的空位 =====
for
(int i = 0; i < rows; i
++) {
for
(int j = 0; j < cols; j
++) {
int num = newresult[i][j];
if
(num >= 0 && num <= 3
) {
vector<pair<
int,
int> > neighbors;
getNeighbors
(i, j, rows, cols, neighbors
);
int light_count = 0;
vector<pair<
int,
int> > available_spots;
for
(int k = 0; k < neighbors.size
(); k
++) {
pair<
int,
int> p = neighbors[k];
int nr = p.first;
int nc = p.second;
int val = newresult[nr][nc];
if
(val == 5
) {
light_count
++;
} else if
(val == 6 && noplace[nr][nc] == 0 && val != 5 && val != 7
) {
available_spots.push_back
(make_pair
(nr, nc
));
}
}
if
(light_count == num
) {
for
(int k = 0; k < available_spots.size
(); k
++) {
pair<
int,
int> p = available_spots[k];
int nr = p.first;
int nc = p.second;
placeLight
(nr, nc, newresult, noplace
);
}
}
}
}
}
// ===== 规则 3（可选）：如果一个空位 6 周围没有灯，就放灯 =====
// 你可以后续加上
}
// 检查是否游戏结束（noplace全为1，且所有数字满足灯数，且灯之间不冲突）
bool ifEnd
(int rows,
int cols, vector<vector<
int> >& newresult, vector<vector<
int> >& noplace
) {
// ===== 第一步：检查 noplace 是否全为 1（即没有可放灯的空位了）=====
for
(int i = 0; i < rows; i
++) {
for
(int j = 0; j < cols; j
++) {
if
(noplace[i][j] == 0&&newresult[i][j]==6
) {
// 只要有一个格子还能放灯（noplace == 0），就不是结束状态
return false;
}
}
}
// ===== 第二步：检查每个数字（0~4）周围 4 方向的灯（5）的数量是否等于该数字 =====
for
(int i = 0; i < rows; i
++) {
for
(int j = 0; j < cols; j
++) {
int val = newresult[i][j];
if
(val >= 0 && val <= 4
) { // 是数字
int expected = val; // 期望的灯数量
int actual = 0;
// 检查上下左右 4 个方向
int dr[] = {-1, 1, 0, 0};
int dc[] = {0, 0, -1, 1};
for
(int d = 0; d < 4; d
++) {
int nr = i
+ dr[d];
int nc = j
+ dc[d];
if
(nr >= 0 && nr < rows && nc >= 0 && nc < cols
) {
if
(newresult[nr][nc] == 5
) { // 是灯
actual
++;
}
}
}
if
(actual != expected
) {
return false; // 数字对应的灯数量不对
}
}
}
}
// ===== 第三步：检查任意两盏灯（5）之间不能无阻挡地看到对方（即不能相互照亮）=====
for
(int i = 0; i < rows; i
++) {
for
(int j = 0; j < cols; j
++) {
if
(newresult[i][j] == 5
) { // 当前是灯
int dr[] = {-1, 1, 0, 0};
int dc[] = {0, 0, -1, 1};
for
(int d = 0; d < 4; d
++) {
int nr = i
+ dr[d];
int nc = j
+ dc[d];
while
(true
) {
if
(nr < 0 || nr >= rows || nc < 0 || nc >= cols
) {
break;
}
int cell = newresult[nr][nc];
if
(cell == 5
) { // 另一盏灯
return false; // 两灯之间无阻挡，冲突！
}
if
(cell == 7 || cell == 0 || cell == 1 || cell == 2 || cell == 3 || cell == 4
) {
// 遇到墙壁或数字，挡住了，不用继续
break;
}
// 如果是空位 6，继续往前检查
nr
+= dr[d];
nc
+= dc[d];
}
}
}
}
}
// 所有条件都满足！
return true;
}
void backtrack
(int k,
int r,
int c,vector<vector<
int> >& newresult, vector<vector<
int> >& noplace
) {
if
(k == r * c
) {
if
(ifEnd
(r, c, newresult, noplace
)) { // ? ifEnd 返回 true 才是满足条件
cout << "? 找到一个合法解！" << endl;
for
(int i = 0; i < r; i
++) {
for
(int j = 0; j < c; j
++) {
cout << newresult[i][j] << " ";
}
cout << endl;
} return;// 如果你只想找一个解，可以在这里 return;
}
else {
for
(int i = 0; i < r; i
++) {
for
(int j = 0; j < c; j
++) {
cout << newresult[i][j] << " ";
}
cout << endl;
}
cout << "? ? 遍历完所有格子，但没有找到满足条件的解。" << endl;
}
}
else if
(noplace[k/c][k%c]==1||newresult[k/c][k%c]==5
){
backtrack
(k
+1,r,c,newresult,noplace
);
}
else {
vector<vector<
int> >tmp1;
vector<vector<
int> >tmp2;
tmp1=newresult;
tmp2=noplace;
placeLight
(k/c,k%c,newresult,noplace
);
canreason
(r, c,newresult,noplace
);
backtrack
(k
+1,r,c,newresult,noplace
);
newresult=tmp1;
noplace=tmp2;
noplace[k/c][k%c]=1;
backtrack
(k
+1,r,c,newresult,noplace
);
}
}
int main
(){
vector<string> result;
int r,c;
cin>>r>>c;
string s;
cin>>s;
result.push_back
(s
);
for
(int j=0;j<result.size
();j
++){
pr
intf
("%s\n",result[j].c_str
());
}
vector<vector<
int> > newresult
(r,vector<
int>
(c
));
vector<vector<
int> > noplace
(r,vector<
int>
(c
));
chushihua
(r,c,result,newresult,noplace
);
backtrack
(0,r,c,newresult,noplace
);
return 0;
}为什么没有回溯的结果，只是遍历完
这是一个Java编程题目，要求在给定的整数范围内判断三个整数A、B和C是否满足A+B大于C的条件。程序通过读取输入的测试用例，运用条件判断避免整数溢出问题，然后输出结果。示例输入和输出展示了不同情况下的判断结果。
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