首页 技术 正文
技术 2022年11月6日
0 收藏 401 点赞 392 浏览 3311 个字

题目链接:时空门问题

简单bfs,每个格子移动的方式除了上下左右,还有时空门,开始想着用邻接表保存每个点能通过时空门到达的点就ok了。很快的敲出来,很快的WA了。长久的dbug并没有发现error。然后换成vector存储,AC,再换成邻接表WA……感觉明明一模一样的好吗…讨厌bug!【怒】

=============================第二天晚上,神奇的大腿发现,我的head数组开成了maxn,然而实际上是maxn*maxn…沃日…最多有maxn*maxn个点啊………………..早知道重构好了….

邻接表代码:AC

#include <stdio.h>
#include <string.h>
#include <iostream>
#define maxn 600
using namespace std;char mp[maxn][maxn];
int n, m;struct Node {
int u, v, nxt;
}edge[maxn*maxn];struct Point {
int x, y;
int step;
}point[maxn*maxn], st, ed, temp;int head[maxn*maxn];
int tot;void addEdge(int u, int v) {
edge[tot].u = u;
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}Point que[maxn*maxn];
int top, tail;
bool vis[maxn][maxn];int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};bool check(Point a) {
if (a.x >= 0 && a.x < n && a.y >= 0 && a.y < m && mp[a.x][a.y] != '#' && !vis[a.x][a.y])
return true;
return false;
}void bfs(Point st) {
top = 0, tail = -1;
vis[st.x][st.y] = 1;
que[++tail] = st; while(top <= tail) {
Point now = que[top++];
if (now.x == ed.x && now.y == ed.y) {
ed.step = now.step;
return;
}
for (int i=0; i<4; ++i) {
Point nxt;
nxt.x = now.x + dir[i][0];
nxt.y = now.y + dir[i][1];
if (check(nxt)) {
vis[nxt.x][nxt.y] = 1;
nxt.step = now.step + 1;
que[++tail] = nxt;
}
} int stp = now.x * m + now.y;
for (int i=head[stp]; i!=-1; i=edge[i].nxt) {
int edp = edge[i].v;
Point nxt;
nxt.x = edp / m, nxt.y = edp % m;
if (check(nxt)) {
vis[nxt.x][nxt.y] = 1;
nxt.step = now.step + 1;
que[++tail] = nxt;
}
}
}
return;
}int main() {
while(~scanf("%d%d", &n, &m)) {
tot = 0;
memset(head, -1, sizeof(head));
memset(vis, 0, sizeof(vis));
getchar();
for (int i=0; i<n; ++i) {
for (int j=0; j<m; ++j) {
scanf("%c", &mp[i][j]);
if (mp[i][j] == 's') {
st.x = i, st.y = j;
st.step = 0;
}
else if (mp[i][j] == 't') {
ed.x = i, ed.y = j;
}
}
if (i != n-1) scanf("\n");
} for (int i=0; i<n*m; ++i) {
int t;
scanf("%d", &t);
for (int j=0; j<t; ++j) {
int edx, edy;
scanf("%d%d", &edx, &edy);
edx -= 1, edy -= 1;
addEdge(i, edx*m+edy);
}
}// for (int i=0; i<n*m; ++i) {
// cout << head[i] << "....\n";
// for (int j=head[i]; j!=-1; j=edge[j].nxt) {
// cout << edge[j].u << " " << edge[j].v << endl;
// }
// cout << "++++++++++++++\n";
// } bfs(st);
printf("%d\n", ed.step);
}
return 0;
}/*
2 3
s.#
..t
1
2 2
0
0
0
1
1 2
0*/

vector代码:AC

#include <stdio.h>
#include <string.h>
#include <iostream>
#define maxn 600
#include <vector>
using namespace std;char mp[maxn][maxn];
int n, m;vector<int>num[maxn*maxn];struct Point {
int x, y;
int step;
}point[maxn*maxn], st, ed, temp;int tot;
Point que[maxn*maxn];
int top, tail;
bool vis[maxn][maxn];int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};bool check(Point a) {
if (a.x >= 0 && a.x < n && a.y >= 0 && a.y < m && mp[a.x][a.y] != '#' && !vis[a.x][a.y])
return true;
return false;
}void bfs(Point st) {
top = 0, tail = -1;
vis[st.x][st.y] = 1;
que[++tail] = st; while(top <= tail) {
Point now = que[top++];
if (now.x == ed.x && now.y == ed.y) {
ed.step = now.step;
return;
}
for (int i=0; i<4; ++i) {
Point nxt;
nxt.x = now.x + dir[i][0];
nxt.y = now.y + dir[i][1];
if (check(nxt)) {
vis[nxt.x][nxt.y] = 1;
nxt.step = now.step + 1;
que[++tail] = nxt;
}
} int stp = now.x * m + now.y;
for (int i=0; i<num[stp].size(); ++i) {
int edp = num[stp][i];
Point nxt;
nxt.x = edp / m, nxt.y = edp % m;
if (check(nxt)) {
vis[nxt.x][nxt.y] = 1;
nxt.step = now.step + 1;
que[++tail] = nxt;
}
}
}
return;
}int main() {
while(~scanf("%d%d", &n, &m)) {
tot = 0;
memset(vis, 0, sizeof(vis));
getchar();
for (int i=0; i<n; ++i) {
for (int j=0; j<m; ++j) {
scanf("%c", &mp[i][j]);
if (mp[i][j] == 's') {
st.x = i, st.y = j;
st.step = 0;
}
else if (mp[i][j] == 't') {
ed.x = i, ed.y = j;
}
num[i*m+j].clear();
}
if (i != n-1) scanf("\n");
} for (int i=0; i<n*m; ++i) {
int t;
scanf("%d", &t);
for (int j=0; j<t; ++j) {
int edx, edy;
scanf("%d%d", &edx, &edy);
edx -= 1, edy -= 1;
num[i].push_back(edx*m+edy);
}
} bfs(st);
printf("%d\n", ed.step);
}
return 0;
}/*
2 3
s.#
..t
1
2 2
0
0
0
1
1 2
0*/

  

上一篇: WebAPI用法
下一篇: hduacm 5104
相关推荐
python开发_常用的python模块及安装方法
adodb:我们领导推荐的数据库连接组件bsddb3:BerkeleyDB的连接组件Cheetah-1.0:我比较喜欢这个版本的cheeta…
日期:2022-11-24 点赞:878 阅读:9,078
Educational Codeforces Round 11 C. Hard Process 二分
C. Hard Process题目连接:http://www.codeforces.com/contest/660/problem/CDes…
日期:2022-11-24 点赞:807 阅读:5,553
下载Ubuntn 17.04 内核源代码
zengkefu@server1:/usr/src$ uname -aLinux server1 4.10.0-19-generic #21…
日期:2022-11-24 点赞:569 阅读:6,402
可用Active Desktop Calendar V7.86 注册码序列号
可用Active Desktop Calendar V7.86 注册码序列号Name: www.greendown.cn Code: &nb…
日期:2022-11-24 点赞:733 阅读:6,177
Android调用系统相机、自定义相机、处理大图片
Android调用系统相机和自定义相机实例本博文主要是介绍了android上使用相机进行拍照并显示的两种方式,并且由于涉及到要把拍到的照片显…
日期:2022-11-24 点赞:512 阅读:7,814
Struts的使用
一、Struts2的获取  Struts的官方网站为:http://struts.apache.org/  下载完Struts2的jar包,…
日期:2022-11-24 点赞:671 阅读:4,898