Mining Station on the Sea
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2584 Accepted Submission(s): 780
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 s2448eabed 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!
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.
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
Sample Output
13
Source
Recommend
题意：有m个海上基站。n个港湾。如今有n仅仅船在n个基站里，基站与基站之间有通讯的船才干够走这条路，告诉基站之间的距离，基站与港湾的距离。如今船要回到港湾，一个港湾仅仅能停靠一仅仅船，并且一旦进去就不能出来了。求全部船都回到港湾要走的最短距离之和。
思路：先用最短路求出每一个船的起始点到每一个港湾的最短距离，而且连边，然后求二分图的最小权匹配。用KM算法。费用流也能够做，但我姿势不够优美超时了。。
。
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 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
")
typedef long long ll;
using namespace std;
const int N=350;
const int MAXM = 1000000;
struct Edge{
int u,v,len,next;
}edge[MAXM];
int n,m,k,p,num;
int dis[N],head[N];
bool inq[N];
int nx,ny; //两边的点数
int g[N][N]; //二分图描写叙述,g赋初值为-INF
int linker[N],lx[N],ly[N]; //y 中各点匹配状态。x，y中的点的标号
int slack[N];
bool visx[N],visy[N];
bool flag;
void init()
{
num=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int len)
{
edge[num]={u,v,len,head[u]};
head[u]=num++;
}
void SPFA(int s)
{
int i,j;
queue<int>Q;
memset(inq,false,sizeof(inq));
memset(dis,INF,sizeof(dis));
Q.push(s);
dis[s]=0;
inq[s]=true;
while (!Q.empty())
{
int u=Q.front();Q.pop();
inq[u]=false;
for (int i=head[u];i+1;i=edge[i].next)
{
int v=edge[i].v;
if (dis[v]>dis[u]+edge[i].len)
{
dis[v]=dis[u]+edge[i].len;
if (!inq[v])
{
Q.push(v);
inq[v]=true;
}
}
}
}
}
bool DFS(int x)
{
visx[x]=true;
for (int y=0;y<ny;y++)
{
if (visy[y]) continue;
int tmp=lx[x]+ly[y]-g[x][y];
if (tmp==0)
{
visy[y]=true;
if (linker[y]==-1||DFS(linker[y]))
{
linker[y]=x;
return true;
}
}
else if (slack[y]>tmp)
slack[y]=tmp;
}
return false;
}
int KM()
{
flag=true;
memset(linker,-1,sizeof(linker));
memset(ly,0,sizeof(ly));
for (int i=0;i<nx;i++) //赋初值。lx置为最大值
{
lx[i]=-INF;
for (int j=0;j<ny;j++)
{
if (g[i][j]>lx[i])
lx[i]=g[i][j];
}
}
for (int x=0;x<nx;x++)
{
for (int i=0;i<ny;i++)
slack[i]=INF;
while (true)
{
memset(visx,false,sizeof(visx));
memset(visy,false,sizeof(visy));
if (DFS(x)) break;
int d=INF;
for (int i=0;i<ny;i++)
if (!visy[i]&&d>slack[i])
d=slack[i];
for (int i=0;i<nx;i++)
if (visx[i])
lx[i]-=d;
for (int i=0;i<ny;i++)
{
if (visy[i])
ly[i]+=d;
else
slack[i]-=d;
}
}
}
int res=0;
for (int i=0;i<ny;i++)
{
if (linker[i]==-1||g[linker[i]][i]<=-INF) //有的点不能匹配的话return-1
{
flag=false;
continue;
}
res+=g[linker[i]][i];
}
return res;
}
//记得nx和ny初始化！！！。！！！
！
int start[N];
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v,cost;
while (~scanf("%d%d%d%d",&n,&m,&k,&p))
{
init();
nx=n;
ny=n;
for (i=0;i<n+m+10;i++)
for (j=0;j<n+m+10;j++)
g[i][j]=-INF;
for (i=1;i<=n;i++)
sf(start[i]);
for (i=0;i<k;i++)
{
sfff(u,v,cost);
addedge(u,v,cost); //网站之间的便能够走多次
addedge(v,u,cost);
}
for (i=0;i<p;i++)
{
sfff(u,v,cost);
addedge(v,u+m,cost); //注意这里是单向边，由于港口仅仅进不出
}
for (i=1;i<=n;i++)
{
SPFA(start[i]); //SPFA求出每一个船起始位置到港湾的最短距离
for (j=1;j<=n;j++)
{
if (dis[j+m]!=INF)
g[i-1][j-1]=-dis[j+m];
// else
// g[i-1][j-1]=0;
}
}
int ans=KM();
printf("%d
",-ans);
}
return 0;
}