POJ-2632的解题思路有哪些？
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本文共计673个文字，预计阅读时间需要3分钟。
伪原创可以改写为：虚假原创。以下是修改后的开头内容：
虚假原创现象在当今网络时代日益突出，这种不尊重知识产权的行为严重扰乱了正常的创作环境。下面将简要探讨虚假原创的成因及影响。
#include <stdio.h>
#include <malloc.h>
struct Robot {
int x;
int y;
char direction;
};
struct Instruction {
int robotId;
char action;
int repeatTime;
};
char direction[4] = {'N', 'E', 'S', 'W'};
char rotate(char originalDirection, char action) {
char returnDirection = 0;
switch (originalDirection) {
case 'N':
if (action == 'L') {
returnDirection = 'W';
} else {
returnDirection = 'E';
}
break;
case 'E':
if (action == 'L') {
returnDirection = 'N';
} else {
returnDirection = 'S';
}
break;
case 'S':
if (action == 'L') {
returnDirection = 'E';
} else {
returnDirection = 'W';
}
break;
case 'W':
if (action == 'L') {
returnDirection = 'S';
} else {
returnDirection = 'N';
}
break;
}
return returnDirection;
}
void forwardRobot(Robot * robot) {
switch (robot->direction) {
case 'N':
robot->y += 1;
break;
case 'E':
robot->x += 1;
break;
case 'S':
robot->y -= 1;
break;
case 'W':
robot->x -= 1;
break;
}
}
void moveRobotAsInstruction(Robot * robot, char action) {
if (action == 'L' || action == 'R') {
robot->direction = rotate(robot->direction, action);
} else if (action == 'F') {
forwardRobot(robot);
}
}
bool isCrashed(Robot * robot, int robotId, int robotNum, int EWdiatance, int NSdistance) {
if ((robot + robotId)->x <= 0 || (robot + robotId)->y <= 0 ||
(robot + robotId)->y > NSdistance || (robot + robotId)->x > EWdiatance) {
printf("Robot %d crashes into the wall\n", robotId + 1);
return true;
}
for (int i = 0; i < robotNum; i++) {
if (i == robotId) {
continue;
}
if ((robot + i)->x == (robot + robotId)->x &&
(robot + i)->y == (robot + robotId)->y) {
printf("Robot %d crashes into robot %d\n", robotId + 1, i + 1);
return true;
}
}
return false;
}
void moveRobots(int EWdiatance, int NSdistance, int robotNum, int instructionNum, Robot * robots, Instruction * instructions) {
for (int i = 0; i < instructionNum; i++) {
for (int j = 0; j < instructions[i].repeatTime; j++) {
// printf("moveRobot %d %c\n", instructions[i].robotId - 1, instructions[i].action);
moveRobotAsInstruction(robots + instructions[i].robotId - 1, instructions[i].action);
if (instructions[i].action != 'L' && instructions[i].action != 'R') {
if (isCrashed(robots, instructions[i].robotId - 1, robotNum,
EWdiatance, NSdistance)) {
return;
}
}
}
}
printf("OK\n");
}
int main() {
int caseNum = 0;
scanf("%d", &caseNum);
for (int i = 0; i < caseNum; i++) {
int EWdiatance;
int NSdistance;
scanf("%d %d", &EWdiatance, &NSdistance);
int robotNum;
int instructionNum;
scanf("%d %d", &robotNum, &instructionNum);
Robot * robots = (Robot *)malloc(robotNum * sizeof(Robot));
Instruction * instructions = (Instruction *)malloc(instructionNum * sizeof(Instruction));
for (int i = 0; i < robotNum; i++) {
scanf("%d %d %c", &(robots[i].x), &(robots[i].y), &(robots[i].direction));
}
for (int i = 0; i < instructionNum; i++) {
scanf("%d %c %d", &(instructions[i].robotId), &(instructions[i].action), &(instructions[i].repeatTime));
// printf("%d %c %d\n", instructions[i].robotId, instructions[i].action, instructions[i].repeatTime);
}
moveRobots(EWdiatance, NSdistance, robotNum, instructionNum, robots, instructions);
free(robots);
robots = NULL;
free(instructions);
instructions = NULL;
}
}
纯模拟， 考察细心和耐心。
有一个技巧就像是转向的处理（不过我懒， 没搞）， 可以搞一个struct 2bit unsigned char成员 存储方向， 在加减越界以后会回到起点或终点，
和转向连转4次回到初始方向一样， 否则写swicth case都巨麻烦。
啥时候自己能做到 纸上写该题code， 一次AC， 就在细心上差不多了。
本文共计673个文字，预计阅读时间需要3分钟。
伪原创可以改写为：虚假原创。以下是修改后的开头内容：
虚假原创现象在当今网络时代日益突出，这种不尊重知识产权的行为严重扰乱了正常的创作环境。下面将简要探讨虚假原创的成因及影响。
#include <stdio.h>
#include <malloc.h>
struct Robot {
int x;
int y;
char direction;
};
struct Instruction {
int robotId;
char action;
int repeatTime;
};
char direction[4] = {'N', 'E', 'S', 'W'};
char rotate(char originalDirection, char action) {
char returnDirection = 0;
switch (originalDirection) {
case 'N':
if (action == 'L') {
returnDirection = 'W';
} else {
returnDirection = 'E';
}
break;
case 'E':
if (action == 'L') {
returnDirection = 'N';
} else {
returnDirection = 'S';
}
break;
case 'S':
if (action == 'L') {
returnDirection = 'E';
} else {
returnDirection = 'W';
}
break;
case 'W':
if (action == 'L') {
returnDirection = 'S';
} else {
returnDirection = 'N';
}
break;
}
return returnDirection;
}
void forwardRobot(Robot * robot) {
switch (robot->direction) {
case 'N':
robot->y += 1;
break;
case 'E':
robot->x += 1;
break;
case 'S':
robot->y -= 1;
break;
case 'W':
robot->x -= 1;
break;
}
}
void moveRobotAsInstruction(Robot * robot, char action) {
if (action == 'L' || action == 'R') {
robot->direction = rotate(robot->direction, action);
} else if (action == 'F') {
forwardRobot(robot);
}
}
bool isCrashed(Robot * robot, int robotId, int robotNum, int EWdiatance, int NSdistance) {
if ((robot + robotId)->x <= 0 || (robot + robotId)->y <= 0 ||
(robot + robotId)->y > NSdistance || (robot + robotId)->x > EWdiatance) {
printf("Robot %d crashes into the wall\n", robotId + 1);
return true;
}
for (int i = 0; i < robotNum; i++) {
if (i == robotId) {
continue;
}
if ((robot + i)->x == (robot + robotId)->x &&
(robot + i)->y == (robot + robotId)->y) {
printf("Robot %d crashes into robot %d\n", robotId + 1, i + 1);
return true;
}
}
return false;
}
void moveRobots(int EWdiatance, int NSdistance, int robotNum, int instructionNum, Robot * robots, Instruction * instructions) {
for (int i = 0; i < instructionNum; i++) {
for (int j = 0; j < instructions[i].repeatTime; j++) {
// printf("moveRobot %d %c\n", instructions[i].robotId - 1, instructions[i].action);
moveRobotAsInstruction(robots + instructions[i].robotId - 1, instructions[i].action);
if (instructions[i].action != 'L' && instructions[i].action != 'R') {
if (isCrashed(robots, instructions[i].robotId - 1, robotNum,
EWdiatance, NSdistance)) {
return;
}
}
}
}
printf("OK\n");
}
int main() {
int caseNum = 0;
scanf("%d", &caseNum);
for (int i = 0; i < caseNum; i++) {
int EWdiatance;
int NSdistance;
scanf("%d %d", &EWdiatance, &NSdistance);
int robotNum;
int instructionNum;
scanf("%d %d", &robotNum, &instructionNum);
Robot * robots = (Robot *)malloc(robotNum * sizeof(Robot));
Instruction * instructions = (Instruction *)malloc(instructionNum * sizeof(Instruction));
for (int i = 0; i < robotNum; i++) {
scanf("%d %d %c", &(robots[i].x), &(robots[i].y), &(robots[i].direction));
}
for (int i = 0; i < instructionNum; i++) {
scanf("%d %c %d", &(instructions[i].robotId), &(instructions[i].action), &(instructions[i].repeatTime));
// printf("%d %c %d\n", instructions[i].robotId, instructions[i].action, instructions[i].repeatTime);
}
moveRobots(EWdiatance, NSdistance, robotNum, instructionNum, robots, instructions);
free(robots);
robots = NULL;
free(instructions);
instructions = NULL;
}
}
纯模拟， 考察细心和耐心。
有一个技巧就像是转向的处理（不过我懒， 没搞）， 可以搞一个struct 2bit unsigned char成员 存储方向， 在加减越界以后会回到起点或终点，
和转向连转4次回到初始方向一样， 否则写swicth case都巨麻烦。
啥时候自己能做到 纸上写该题code， 一次AC， 就在细心上差不多了。