Mining Station on the Sea
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2997 Accepted Submission(s): 913
Problem Description
The
ocean is a treasure house of resources and the development of human
society comes to depend more and more on it. In order to develop and
utilize marine resources, it is necessary to build mining stations on
the sea. However, due to seabed mineral resources, the radio signal in
the sea is often so weak that not all the mining stations can carry out
direct communication. However communication is indispensable, every two
mining stations must be able to communicate with each other (either
directly or through other one or more mining stations). To meet the need
of transporting the exploited resources up to the land to get put into
use, there build n ports correspondently along the coast and every port
can communicate with one or more mining stations directly.
ocean is a treasure house of resources and the development of human
society comes to depend more and more on it. In order to develop and
utilize marine resources, it is necessary to build mining stations on
the sea. However, due to seabed mineral resources, the radio signal in
the sea is often so weak that not all the mining stations can carry out
direct communication. However communication is indispensable, every two
mining stations must be able to communicate with each other (either
directly or through other one or more mining stations). To meet the need
of transporting the exploited resources up to the land to get put into
use, there build n ports correspondently along the coast and every port
can communicate with one or more mining stations directly.
Due to
the fact that some mining stations can not communicate with each other
directly, for the safety of the navigation for ships, ships are only
allowed to sail between mining stations which can communicate with each
other directly.
The mining is arduous and people do this job
need proper rest (that is, to allow the ship to return to the port). But
what a coincidence! This time, n vessels for mining take their turns to
take a rest at the same time. They are scattered in different stations
and now they have to go back to the port, in addition, a port can only
accommodate one vessel. Now all the vessels will start to return, how to
choose their navigation routes to make the total sum of their sailing
routes minimal.
Notice that once the ship entered the port, it will not come out!
Input
There
are several test cases. Every test case begins with four integers in
one line, n (1 = <n <= 100), m (n <= m <= 200), k and p. n
indicates n vessels and n ports, m indicates m mining stations, k
indicates k edges, each edge corresponding to the link between a mining
station and another one, p indicates p edges, each edge indicating the
link between a port and a mining station. The following line is n
integers, each one indicating one station that one vessel belongs to.
Then there follows k lines, each line including 3 integers a, b and c,
indicating the fact that there exists direct communication between
mining stations a and b and the distance between them is c. Finally,
there follows another p lines, each line including 3 integers d, e and
f, indicating the fact that there exists direct communication between
port d and mining station e and the distance between them is f. In
addition, mining stations are represented by numbers from 1 to m, and
ports 1 to n. Input is terminated by end of file.
are several test cases. Every test case begins with four integers in
one line, n (1 = <n <= 100), m (n <= m <= 200), k and p. n
indicates n vessels and n ports, m indicates m mining stations, k
indicates k edges, each edge corresponding to the link between a mining
station and another one, p indicates p edges, each edge indicating the
link between a port and a mining station. The following line is n
integers, each one indicating one station that one vessel belongs to.
Then there follows k lines, each line including 3 integers a, b and c,
indicating the fact that there exists direct communication between
mining stations a and b and the distance between them is c. Finally,
there follows another p lines, each line including 3 integers d, e and
f, indicating the fact that there exists direct communication between
port d and mining station e and the distance between them is f. In
addition, mining stations are represented by numbers from 1 to m, and
ports 1 to n. Input is terminated by end of file.
Output
Each test case outputs the minimal total sum of their sailing routes.
Sample Input
3 5 5 6
1 2 4
1 3 3
1 4 4
1 5 5
2 5 3
2 4 3
1 1 5
1 5 3
2 5 3
2 4 6
3 1 4
3 2 2
1 2 4
1 3 3
1 4 4
1 5 5
2 5 3
2 4 3
1 1 5
1 5 3
2 5 3
2 4 6
3 1 4
3 2 2
Sample Output
13
题意:现在有m个油井和n个港口(n<=m),现在有n条船停在这些油井这里,第一行输入n个数, 输入IDX[i]代表第i条船停在 IDX[i]这个油井这里,然后接下来有k行,输入u，v，w代表油井u和油井v之间的距离为w,然后又p行,代表了港口和油井之间的距离,现在这些船全部要回到港口,而且每个港口只能停一艘船,问这些船返回港口的最短距离。
题解:开始的时候建边,要为油田和油田建双向边，但是油田港口只能建单向边,因为回去了就不能往回走了.用spfa求出每艘船到每个港口的距离(注意编号,港口编号为 1+m~n+m)，然后求出来之后进行KM算法最优匹配即可得到答案。
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
using namespace std;
const int N = ;
const int INF = ;
int lx[N],ly[N];
bool visitx[N],visity[N];
int slack[N];
int match[N];
int graph[N][N];
int idx[N]; ///船所在的油井下标
int n,m,k,p;
bool Hungary(int u)
{
visitx[u] = true;
for(int i=m+; i<=m+n; i++)
{
if(!visity[i])
{
int temp = lx[u]+ly[i]-graph[u][i];
if(temp==)
{
visity[i] = true;
if(match[i]==-||Hungary(match[i]))
{
match[i] = u;
return true;
}
}
else
{
slack[i] = min(slack[i],temp);
}
}
}
return false;
}
void KM()
{
memset(match,-,sizeof(match));
memset(ly,,sizeof(ly));
for(int i=; i<=n; i++) ///定标初始化
{
lx[idx[i]] = -INF;
}
for(int i=; i<=n; i++)
{
for(int j=m+; j<=n+m; j++)
{
lx[idx[i]] = max(lx[idx[i]],graph[idx[i]][j]);
}
}
for(int i=; i<=n; i++)
{
for(int j=m+; j<=m+n; j++) slack[j] = INF;
while(true)
{
memset(visitx,false,sizeof(visitx));
memset(visity,false,sizeof(visity));
if(Hungary(idx[i])) break;
else
{
int temp = INF;
for(int j=+m; j<=n+m; j++)
{
if(!visity[j]) temp = min(temp,slack[j]);
}
for(int j=; j<=n; j++)
{
if(visitx[idx[j]]) lx[idx[j]]-=temp;
}
for(int j=; j<=n+m; j++)
{
if(visity[j]) ly[j]+=temp;
else slack[j]-=temp;
}
}
}
}
}
struct Edge
{
int v,w,next;
} edge[N*N];
int head[N],tot;
void init()
{
memset(head,-,sizeof(head));
tot = ;
}
void addEdge(int u,int v,int w,int &k)
{
edge[k].v = v,edge[k].w = w,edge[k].next = head[u],head[u] = k++;
}
int low[N];
bool vis[N];
void spfa(int s)
{
for(int i=; i<=n+m; i++)
{
low[i] = INF;
vis[i] = false;
}
queue<int> q;
low[s] = ;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i=head[u]; i!=-; i=edge[i].next)
{
int v = edge[i].v,w = edge[i].w;
if(low[v]>low[u]+w)
{
low[v] = low[u]+w;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
for(int i=+m; i<=n+m; i++)
{
if(low[i]!=INF)
{
graph[s][i] = -low[i];
}
}
}
int main()
{
while(scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF)
{
init();
for(int i=; i<=n; i++)
{
scanf("%d",&idx[i]);
}
/**油田 1-m,港口 m+1-m+n*/
for(int i=; i<=k; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addEdge(u,v,w,tot);
addEdge(v,u,w,tot);
}
for(int i=; i<=p; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addEdge(v,u+m,w,tot); ///油井向港口添加单向边
}
memset(graph,,sizeof(graph));
for(int i=; i<=n; i++)
{
spfa(idx[i]);
}
KM();
int ans = ;
for(int i=+m; i<=n+m; i++)
{
if(match[i]!=-)
{
ans+=graph[match[i]][i];
}
}
printf("%d\n",-ans);
}
}