hdu3468 最大流/二分匹配+BFS
Treasure Hunting
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 903 Accepted Submission(s): 228
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?
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.
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
Recommend
zhouzeyong
参考了别人的思路~~
网络流建图是难点啊！！
http://www.cnblogs.com/woaishizhan/archive/2013/04/08/3008719.html
http://blog.csdn.net/wall_f/article/details/8990937
#include <cstring>
#include <cstdio>
#include <queue>
#include <cstdlib>
#include <cctype>
#define MAXN 105*105*2
#define MAXM 105*105*100
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
int u,v,f;
};
node e[MAXM];
int first[MAXN],next[MAXM];
int gap[MAXN],d[MAXN],curedge[MAXN],pre[MAXN];
int cc, n,m;
inline void add_edge(int u,int v,int f)
{
e[cc].u=u;
e[cc].v=v;
e[cc].f=f;
next[cc]=first[u];
first[u]=cc;
cc++;
e[cc].u=v;
e[cc].v=u;
e[cc].f=0;
next[cc]=first[v];
first[v]=cc;
cc++;
}
int ISAP(int s,int t,int n)
{
int cur_flow,flow_ans=0,u,tmp,neck,i,v;
memset(d,0,sizeof(d));
memset(gap,0,sizeof(gap));
memset(pre,-1,sizeof(pre));
for(i=0;i<=n;i++)
curedge[i]=first[i];
gap[0]=n+1;
u=s;
while(d[s]<=n)
{
if(u==t)
{
cur_flow=inf;
for(i=s;i!=t;i=e[curedge[i]].v)
{
if(cur_flow>e[curedge[i]].f)
{
neck=i;
cur_flow=e[curedge[i]].f;
}
}
for(i=s;i!=t;i=e[curedge[i]].v)
{
tmp=curedge[i];
e[tmp].f-=cur_flow;
e[tmp^1].f+=cur_flow;
}
flow_ans+=cur_flow;
u=neck;
}
for(i=curedge[u];i!=-1;i=next[i])
{
v=e[i].v;
if(e[i].f&&d[u]==d[v]+1)
break;
}
if(i!=-1)
{
curedge[u]=i;
pre[v]=u;
u=v;
}
else
{
if(0==--gap[d[u]])
break;
curedge[u]=first[u];
for(tmp=n+5,i=first[u];i!=-1;i=next[i])
if(e[i].f)
tmp=min(tmp,d[e[i].v]);
d[u]=tmp+1;
++gap[d[u]];
if(u!=s)
u=pre[u];
}
}
return flow_ans;
}
char map[105][105];
int rally[105*105];
int gold[105*105];
int pos[105*105];
int dist[55][105*105];
int vis[105][105];
int xx[4]={0,1,0,-1};
int yy[4]={1,0,-1,0};
int trans(char word)
{
if(word>='A'&&word<='Z')
return word-'A';
else
return word-'a'+26;
}
void bfs(int s)
{
queue<int> q;
memset(vis,0,sizeof(vis));
memset(dist[s],inf,sizeof(dist[s]));
int t=pos[s];
vis[t/m][t%m]=1;
q.push(t);
dist[s][t]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
int x=u/m;
int y=u%m;
int i;
for(i=0;i<4;i++)
{
int nx=x+xx[i];
int ny=y+yy[i];
if(nx<0||nx>=n||ny<0||ny>=m)
continue;
if(vis[nx][ny]||map[nx][ny]=='#')
continue;
vis[nx][ny]=1;
dist[s][nx*m+ny]=dist[s][x*m+y]+1;
q.push(nx*m+ny);
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(first,-1,sizeof(first));
memset(next,-1,sizeof(next));
cc=0;
memset(pos,0,sizeof(pos));
int i,j;
int cnt1=0;
int cnt2=0;
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
for(j=0;j<m;j++)
{
if(isalpha(map[i][j]))
{
pos[trans(map[i][j])]=i*m+j;
cnt1++;
}
else if(map[i][j]=='*')
{
gold[cnt2]=i*m+j;
cnt2++;
}
}
}
for(i=0;i<cnt1;i++)
bfs(i);
int flag=1;
for(i=1;i<=cnt1-1;i++)
{
for(j=0;j<cnt2;j++)
{
if(dist[i][gold[j]]+dist[i-1][gold[j]]==dist[i-1][pos[i]])
{
add_edge(i,cnt1+j,1);
}
if(dist[i-1][pos[i]]==inf)
{
flag=0;
}
}
}
if(!flag)
{
printf("-1\n");
continue;
}
int s=0,t=cnt1+cnt2;
for(i=1;i<=cnt1-1;i++)
add_edge(s,i,1);
for(i=0;i<cnt2;i++)
add_edge(cnt1+i,t,1);
int res=ISAP(s,t,t);
printf("%d\n",res);
}
return 0;
}