Croc Champ 2012 - Round 1 B. Chamber of Secrets(二分图 最短路)
2017-07-11 20:02
323 查看
题意: 一个n*m的地图(n,m <=1e3),有一些镜子,透过镜子你能往上下左右看,现在你在(1, 0)位置往右看,问最少透过几块镜子可以看到(n, m+1)处的东西。
思路:比较巧妙的建图。要走最少步走到右下角,想到的肯定是最短路,但是怎么建图呢?
把每个点(x,y)拆成两个点, 横坐标看作x点,纵坐标看作y+n点,每个'#;建x -> y+n和y+n -> x的双向边。这样就成了个二分图,每次可以从横坐标到纵坐标或是
纵坐标到横坐标,横坐标1是起点,横坐标n是终点,跑一下最短路就行了。
代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = (1e3+5)*2;
const int INF = 0x3f3f3f3f;
char str[maxn][maxn];
int n, m, dis[maxn], vis[maxn];
vector<int> g[maxn];
int spfa()
{
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
queue<int> q;
dis[1] = 0;
q.push(1);
while(!q.empty())
{
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = 0; i < g[u].size(); i++)
{
int v = g[u][i];
if(dis[u]+1 < dis[v])
{
dis[v] = dis[u]+1;
if(!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
return dis
==INF ? -1 : dis
;
}
int main(void)
{
while(cin >> n >> m)
{
for(int i = 0; i < maxn; i++)
g[i].clear();
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
char ch;
scanf(" %c", &ch);
if(ch == '#')
{
g[i].push_back(j+n);
g[j+n].push_back(i);
}
}
printf("%d\n", spfa());
}
return 0;
}
思路:比较巧妙的建图。要走最少步走到右下角,想到的肯定是最短路,但是怎么建图呢?
把每个点(x,y)拆成两个点, 横坐标看作x点,纵坐标看作y+n点,每个'#;建x -> y+n和y+n -> x的双向边。这样就成了个二分图,每次可以从横坐标到纵坐标或是
纵坐标到横坐标,横坐标1是起点,横坐标n是终点,跑一下最短路就行了。
代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = (1e3+5)*2;
const int INF = 0x3f3f3f3f;
char str[maxn][maxn];
int n, m, dis[maxn], vis[maxn];
vector<int> g[maxn];
int spfa()
{
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
queue<int> q;
dis[1] = 0;
q.push(1);
while(!q.empty())
{
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = 0; i < g[u].size(); i++)
{
int v = g[u][i];
if(dis[u]+1 < dis[v])
{
dis[v] = dis[u]+1;
if(!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
return dis
==INF ? -1 : dis
;
}
int main(void)
{
while(cin >> n >> m)
{
for(int i = 0; i < maxn; i++)
g[i].clear();
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
char ch;
scanf(" %c", &ch);
if(ch == '#')
{
g[i].push_back(j+n);
g[j+n].push_back(i);
}
}
printf("%d\n", spfa());
}
return 0;
}
相关文章推荐
- Codeforces Croc Champ 2012 - Round 2 B. Word Cut
- cf Croc Champ 2012 - Round 1 C. Spiral Maximum
- CF_Croc Champ 2012 - Round 2 (Unofficial Div. 2 Edition)
- Codeforces Croc Champ 2012 - Round 2 (Unofficial Div. 2 Edition)
- Codeforces Croc Champ 2012 - Round 2 (Unofficial Div. 2 Edition)
- CF 173B Chamber of Secrets 最短路
- Croc Champ 2013 - Round 2 Div. 2 D
- poj2942 Knights of the Round Table,无向图点双联通,二分图判定
- CROC-MBTU 2012, Elimination Round / 245D Restoring Table (位运算)
- UVALive3523-Knights of the Round Table(BCC+二分图判定)
- CF CROC-MBTU 2012, Final Round div 2
- 【连通图|双连通+二分图判定】POJ-2942 Knights of the Round Table
- CROC-MBTU 2012, Elimination Round / 245E Mishap in Club (想法题)
- poj 2942 Knights of the Round Table 补图+点双连通分量+判定二分图
- Croc Champ 2013 - Round 2 (Div. 2 Edition) C
- Croc Champ 2013 - Round 2 题解
- UVALive - 3523 Knights of the Round Table(【点双连通分量】+【二分图判定】)
- poj 2942 Knights of the Round Table(无向图的双连通分量+二分图判定)
- Croc Champ 2013 - Round 2 题解
- Croc Champ 2013 - Round 1 E. Copying Data(线段树)