Treasure Hunting
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1509 Accepted Submission(s): 393
Problem Description
Do you like treasure hunting? Today, with one of his friend, iSea is on a venture trip again. As most movie said, they find so many gold hiding in their trip.
Now iSea’s clever friend has already got the map of the place they are going to hunt, simplify the map, there are three ground types:
● ‘.‘ means blank ground, they can get through it
● ‘#‘ means block, they can’t get through it
● ‘*‘ means gold hiding under ground, also they can just get through it (but you won’t, right?)
What makes iSea very delighted is the friend with him is extraordinary justice, he would not take away things which doesn’t belong to him, so all the treasure belong to iSea oneself!
But his friend has a request, he will set up a number of rally points on the map, namely ‘A‘, ‘B‘ ... ‘Z‘, ‘a‘, ‘b‘ ... ‘z‘ (in that order, but may be less than 52), they start in ‘A‘, each time friend reaches to the next rally point in the shortest way, they
have to meet here (i.e. iSea reaches there earlier than or same as his friend), then start together, but you can choose different paths. Initially, iSea’s speed is the same with his friend, but to grab treasures, he save one time unit among each part of road,
he use the only one unit to get a treasure, after being picked, the treasure’s point change into blank ground.
Under the premise of his friend’s rule, how much treasure iSea can get at most?
Input
There are several test cases in the input.
Each test case begin with two integers R, C (2 ≤ R, C ≤ 100), indicating the row number and the column number.
Then R strings follow, each string has C characters (must be ‘A’ – ‘Z’ or ‘a’ – ‘z’ or ‘.’ or ‘#’ or ‘*’), indicating the type in the coordinate.
The input terminates by end of file marker.
Output
For each test case, output one integer, indicating maximum gold number iSea can get, if they can’t meet at one or more rally points, just output -1.
Sample Input
2 4 A.B. ***C 2 4 A#B. ***C
Sample Output
1 2
Author
iSea @ WHU
Source
2010 ACM-ICPC Multi-University
Training Contest（3）——Host by WHU
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题意：在一个R*C的地图内，字母表示集合点，‘*’表示宝藏，‘.’表示空地，现在沿着A->....->Z->a->....->z的方向走，途中从一个集合点到下一个集合点之间只能捡一个宝藏，问最后最多能捡多少宝藏。
思路：将集合点和宝藏分别看成两个集合，若在集合点x到y的最短路径上有‘*’，那么就在x和‘*’之间连边。bfs求出所有集合点到下一个集合点的最短路径。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Node
{
int x,y;
int step;
};
int dir[4][2]={1,0,-1,0,0,1,0,-1};
int g[60][10005];
int linker[10005];
char mp[105][105];
int id1[105][105],id2[105][105];
int dist1[60][10005],dist2[60][60];
int R,C,uN,vN;
bool used[10005];
bool vis[105][105];
bool dfs(int u)
{
for (int v=1;v<=vN;v++)
{
if (g[u][v]&&!used[v])
{
used[v]=true;
if (linker[v]==-1||dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
}
return false;
}
int hungary()
{
int res=0;
memset(linker,-1,sizeof(linker));
for (int u=1;u<=uN;u++)
{
memset(used,false,sizeof(used));
if (dfs(u)) res++;
}
return res;
}
bool isok(Node a)
{
if (a.x>=0&&a.x<R&&a.y>=0&&a.y<C) return true;
return false;
}
void bfs(int sx,int sy)
{
int minn=INF;
queue<Node>Q;
Node st,now;
while (!Q.empty()) Q.pop();
st.x=sx;st.y=sy;
st.step=0;
memset(vis,false,sizeof(vis));
vis[sx][sy]=true;
Q.push(st);
// printf("%d %d+\n",sx,sy);
while (!Q.empty())
{
st=Q.front(); Q.pop();
for (int i=0;i<4;i++)
{
now.x=st.x+dir[i][0];
now.y=st.y+dir[i][1];
now.step=st.step+1;
if (isok(now)&&mp[now.x][now.y]!='#'&&!vis[now.x][now.y])
{
// printf("%d %d\n",now.x,now.y);
vis[now.x][now.y]=true;
Q.push(now);
if (mp[now.x][now.y]=='*')
dist1[id1[sx][sy]][id2[now.x][now.y]]=now.step;
if (id1[now.x][now.y]==id1[sx][sy]+1)
dist2[id1[sx][sy]][id1[now.x][now.y]]=now.step;
}
}
}
}
int main()
{
// freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
int i,j;
while (~scanf("%d%d",&R,&C))
{
int cnt1=0,cnt2=1;
int have[60];
memset(have,0,sizeof(have));
memset(id1,0,sizeof(id1));
memset(g,0,sizeof(g));
memset(id2,0,sizeof(id2));
for (i=0;i<R;i++)
{
scanf("%s",mp[i]);
for (j=0;j<C;j++)
{
if (mp[i][j]>='A'&&mp[i][j]<='Z')
{
id1[i][j]=mp[i][j]-'A'+1; //给集合点标号
cnt1++;
have[id1[i][j]]=1;
}
else if (mp[i][j]>='a'&&mp[i][j]<='z')
{
id1[i][j]=mp[i][j]-'a'+27; //给集合点标号
cnt1++;
have[id1[i][j]]=1;
}
else if (mp[i][j]=='*')
id2[i][j]=cnt2++; //给宝藏标号
}
}
for (i=1;i<=cnt1;i++) //若集合点中间断层了直接输出-1
if (!have[i]) break;
if (i<=cnt1){
printf("-1\n");
continue;
}
for (i=0;i<cnt1+10;i++)
for (j=0;j<cnt1+10;j++)
dist2[i][j]=-1; //dist2[i][[j]表示集合点i到集合点j的最短距离
for (i=0;i<cnt1+10;i++)
for (j=0;j<cnt2+10;j++)
dist1[i][j]=-1; //dist1[i][j]表示集合点i到宝藏j的最短距离
int flag=1;
for (i=0;i<R;i++)
{
for (j=0;j<C;j++)
{
if (id1[i][j])
bfs(i,j);
}
}
for (i=1;i<cnt1;i++) //若有两个集合点之间不可到达直接输出-1
if (dist2[i][i+1]==-1){
flag=0;
break;
}
if (!flag)
{
// DBG;
printf("-1\n");
continue;
}
// DBG;
for (i=0;i<R;i++)//建图
{
for (j=0;j<C;j++)
{
if (id1[i][j])
{
for (int k=1;k<cnt2;k++)
{
if (dist1[id1[i][j]][k]+dist1[id1[i][j]+1][k]==dist2[id1[i][j]][id1[i][j]+1])
g[id1[i][j]][k]=1;
}
}
}
}
uN=cnt1;vN=cnt2-1;
printf("%d\n",hungary());
}
return 0;
}