Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22390 Accepted Submission(s): 7304
Problem Description
Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
Input
The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF
Output
For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
Sample Input
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
...#
Sample Output
66
88
66
Author
yifenfei
【分析】：dis开了个三维数组，第三维表示不同的两次搜索，每次给第三维传递一个参数来表示哪一次。另外对于dis初始化的时候应初始化无穷大，因为有可能某个KFC根本就无法到达，影响最后求结果。因为@（终点）有多个，需要用数组来保存到每个点的最短路，用0、1区分两个不同起点。之前数组开小，显示的却是TLE，搞不懂HDU= =
【代码】：
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 500; //数组之前开小了...但是一直显示TLE而不是RE？？？我爆哭
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0};
const int dy[] = {0,0,1,-1};
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n,m,f;
char a[maxn][maxn];
int d[maxn][maxn][2];
int v[maxn][maxn];
struct node
{
int x,y,step;
}st,ed;
bool ok(int x,int y)
{
return x>=0 && y>=0 && x<n && y<m && !v[x][y] && a[x][y]!='#';
}
void bfs(int x,int y,int f)
{
queue<node> q;
ms(v,0); //每搜索一次之前都要清空vis
v[x][y]=1;
st.x=x,st.y=y,st.step=0;
q.push(st);
while(!q.empty())
{
st = q.front();
q.pop();
if(a[st.x][st.y]=='@')
d[st.x][st.y][f]=st.step;
for(int i=0;i<4;i++)
{
ed.x = st.x + dir[i][0];
ed.y = st.y + dir[i][1];
if(!ok(ed.x,ed.y)) continue;
ed.step = st.step + 1;
v[ed.x][ed.y] = 1;
q.push(ed);
}
}
}
int main()
{
//需要用数组保存到每个点的最短路，因为这里需要多次查询路径
while(~scanf("%d%d",&n,&m))
{
int ans=INF;
int x1,x2,y1,y2;
getchar();
rep(i,0,n)
{
scanf("%s",&a[i]);
}
rep(i,0,n)
{
rep(j,0,m)
{
if(a[i][j]=='Y')
{
x1=i,y1=j;
}
else if(a[i][j]=='M')
{
x2=i,y2=j;
}
}
}
ms(d,INF);
bfs(x1,y1,0);
bfs(x2,y2,1);
rep(i,0,n)
{
rep(j,0,m)
{
if(a[i][j]=='@') //对于多个KFC地点，求出两个人同时到达某一个KFC的最小步数的和，取最小值就可以了
{
ans = min(ans,d[i][j][0] + d[i][j][1]);
}
}
}
printf("%d
",ans*11);
}
}
/*
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#
*/