Uva - 810 - A Dicey Problem
根据状态进行bfs,手动打表维护骰子滚动。
AC代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset>
#include <cassert>
#include <cmath>
using namespace std;
const int maxn = 65536;
const int dx[] = { -1, 1, 0, 0 };
const int dy[] = { 0, 0, -1, 1 };
struct State {
int x, y, top, front;
State *prev;
} states[maxn];
int statesIdx;
int n, m, sx, sy, dtop, dfront;
int g[32][32];
State* getNewState(int x, int y, int dtop, int dfront, State *prev = NULL) {
State *p = &states[statesIdx++];
assert(statesIdx < maxn);
p->x = x, p->y = y, p->top = dtop, p->front = dfront, p->prev = prev;
return p;
}
int diceTable[7][7]; // [front][top] = right
void rotateDice(int dir, int dtop, int dfront, int &rtop, int &rfront) {
rtop = rfront = 1;
if (dir == 0) { // 向上滚动
rtop = dfront, rfront = 7 - dtop;
}
else if (dir == 1) { // 向下滚动
rtop = 7 - dfront, rfront = dtop;
}
else if (dir == 2) { // 向左滚动
rfront = dfront;
rtop = diceTable[dfront][dtop];
}
else if (dir == 3) { // 向右滚动
rfront = dfront;
rtop = 7 - diceTable[dfront][dtop];
}
}
void bfs(int sx, int sy, int dtop, int dfront) {
statesIdx = 0;
int used[32][32][7][7] = {}; // used[x][y][dtop][dfront]
int tx, ty, tdtop, tdfront;
queue<State*> Q;
State *u, *v;
Q.push(getNewState(sx, sy, dtop, dfront));
used[sx][sy][dtop][dfront] = 1;
while (!Q.empty()) {
u = Q.front(), Q.pop();
for (int i = 0; i < 4; i++) {
tx = u->x + dx[i], ty = u->y + dy[i];
if (tx <= 0 || ty <= 0 || tx > n || ty > m) {
continue;
}
if (g[tx][ty] != -1 && g[tx][ty] != u->top) {
continue;
}
if (tx == sx && ty == sy) {
vector< pair<int, int> > ret;
State *p = u;
ret.push_back(make_pair(sx, sy));
while (p != NULL) {
ret.push_back(make_pair(p->x, p->y));
p = p->prev;
}
for (int i = ret.size() - 1, j = 0; i >= 0; i--, j++) {
if (j % 9 == 0) printf(" ");
if (j % 9 != 0) printf(",");
printf("(%d,%d)", ret[i].first, ret[i].second);
if (j % 9 == 8 || i == 0) {
if (i) {
printf(",\n");
}
else {
puts("");
}
}
}
return;
}
rotateDice(i, u->top, u->front, tdtop, tdfront);
if (used[tx][ty][tdtop][tdfront]) {
continue;
}
used[tx][ty][tdtop][tdfront] = 1;
v = getNewState(tx, ty, tdtop, tdfront, u);
Q.push(v);
}
}
printf(" No Solution Possible\n");
}
int main() {
// 手动打表维护骰子滚动
// diceface[front][top] = right
diceTable[1][2] = 4, diceTable[1][3] = 2, diceTable[1][4] = 5, diceTable[1][5] = 3;
diceTable[2][1] = 3, diceTable[2][3] = 6, diceTable[2][4] = 1, diceTable[2][6] = 4;
diceTable[3][1] = 5, diceTable[3][2] = 1, diceTable[3][5] = 6, diceTable[3][6] = 2;
diceTable[4][1] = 2, diceTable[4][2] = 6, diceTable[4][5] = 1, diceTable[4][6] = 5;
diceTable[5][1] = 4, diceTable[5][3] = 1, diceTable[5][4] = 6, diceTable[5][6] = 3;
diceTable[6][2] = 3, diceTable[6][3] = 5, diceTable[6][4] = 2, diceTable[6][5] = 4;
char testcase[1024];
while (scanf("%s", testcase) == 1) {
if (!strcmp("END", testcase))
break;
scanf("%d %d %d %d %d %d", &n, &m, &sx, &sy, &dtop, &dfront);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%d", &g[i][j]);
}
}
printf("%s\n", testcase);
bfs(sx, sy, dtop, dfront);
}
return 0;
}
Uva - 810 - A Dicey Problem的更多相关文章
- UVA 810 - A Dicey Problem(BFS)
UVA 810 - A Dicey Problem 题目链接 题意:一个骰子,给你顶面和前面.在一个起点,每次能移动到周围4格,为-1,或顶面和该位置数字一样,那么问题来了,骰子能不能走一圈回到原地, ...
- UVA 810 A Dicey Promblem 筛子难题 (暴力BFS+状态处理)
读懂题意以后还很容易做的, 和AbbottsRevenge类似加一个维度,筛子的形态,可以用上方的点数u和前面的点数f来表示,相对的面点数之和为7,可以预先存储u和f的对应右边的点数,点数转化就很容易 ...
- UVa 101 The Blocks Problem Vector基本操作
UVa 101 The Blocks Problem 一道纯模拟题 The Problem The problem is to parse a series of commands that inst ...
- 【暑假】[深入动态规划]UVa 1380 A Scheduling Problem
UVa 1380 A Scheduling Problem 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=41557 ...
- UVA - 524 Prime Ring Problem(dfs回溯法)
UVA - 524 Prime Ring Problem Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & % ...
- uva 10837 - A Research Problem(欧拉功能+暴力)
题目链接:uva 10837 - A Research Problem 题目大意:给定一个phin.要求一个最小的n.欧拉函数n等于phin 解题思路:欧拉函数性质有,p为素数的话有phip=p−1; ...
- poj 1872 A Dicey Problem WA的代码,望各位指教!!!
A Dicey Problem Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 832 Accepted: 278 Des ...
- poj1872A Dicey Problem
Home Problems Status Contest 284:28:39 307:00:00 Overview Problem Status Rank A B C D E F G H ...
- UVA 10026 Shoemaker's Problem 鞋匠的难题 贪心+排序
题意:鞋匠一口气接到了不少生意,但是做鞋需要时间,鞋匠只能一双一双地做,根据协议每笔生意如果拖延了要罚钱. 给出每笔生意需要的天数和每天的罚钱数,求出最小罚钱的排列顺序. 只要按罚款/天数去从大到小排 ...
随机推荐
- 2. struct A 和 typedef struct A
2. struct A 和 typedef struct A 2.1 struct A struct A{}定义一个名为struct A的结构体. 下例定义了struct A同时,声明了两个变量(注意 ...
- VueJs(4)---V-model指令
V-model指令 摘要 限制: v-model只能用在:<input> <select> <textarea> <components&g ...
- c++ 深入理解虚函数
为什么使用虚函数?什么是虚函数?虚函数是为了解决什么问题? 面向对象的三大特征: 封装 多态 继承 普通虚函数 虚析构函数 纯虚函数 抽象类 接口类 隐藏 vs 覆盖 隐藏与覆盖之间的关系 早绑定和晚 ...
- 继承自 DevExpress 17.2 的自定义控件如何在工具箱显示
最近把DevExpress版本从13.1升级到了17.2,结果发现继承自DevExpress的自定义控件居然在工具箱中消失了,弄了两天还是没有任何头绪,部分自定义Dev控件可以正常出现,但大部分自定义 ...
- file的基本操作;file的修改
file的基本操作 # Author:nadech # 文件读写/修改/ #data = open("yesterday",encoding="utf-8"). ...
- E1
en表"使怎么样" engage 吸引,从事,订婚 be engaged in doing sth. 忙于 endure 忍耐,忍受 enforce 强制执行 enrol ...
- 没有JavaScript的基础,我可以学习Angular2吗?
Can I learn and understand Angular2 without understanding JavaScript? 没有JavaScript基础我能学习和理解Angular2吗 ...
- CentOS 7 下使用虚拟环境Virtualenv安装Tensorflow cpu版记录
1.首先安装pip-install 在使用centos7的软件包管理程序yum安装python-pip的时候会报一下错误: No package python-pip available. Error ...
- Compile C++ code in Matlab with OpenCV support
Provides a function named as "mex_opencv(src)" The code function mex_opencv(src) ARC = 'x6 ...
- post插件
分享牛系列,分享牛专栏,分享牛.在项目开发中,http请求方式是最常见的了.怎么模拟http请求呢?方法有很多种,可以使用httpclient直接模拟请求,也可以使用火狐post插件方式,这个章节主要 ...