Codeforces Gym - 101635G [KM]
2017-12-19 18:00
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题意:有n个酒,m个快递员,一个酒店,目的是将所有的酒运到酒店里,每个快递员每一次只能送一瓶酒,并送到酒店,问需要的最小总距离是多少。
题解:由于快递员到了酒店之后所有的快递员其实都是一样的了,所以我们建立n-1个虚拟点表示必须有一个快递员进入酒店然后可以继续不断地运送,之后建立n*(n+m-1)条运送路线,设立负边权,然后跑KM得到最小匹配权值。
AC代码:
struct node
{
int x, y;
node(){}
node(int xx,int yy)
{
this->x = xx;
this->y = yy;
}
};
int dis(node x, node y)
{
return abs(x.x - y.x) + abs(x.y - y.y);
}
int main()
{
int n, m,x,y;
scanf("%d%d", &n, &m);
memset(g, 0, sizeof(g));
vector<node>v1,v2;
for (int i=0;i<n;i++){
scanf("%d%d",&x,&y);
v1.push_back(node(x,y));
}
for (int i=0;i<m;i++){
scanf("%d%d",&x,&y);
v2.push_back(node(x,y));
}
node p;
scanf("%d%d",&p.x,&p.y);
int ans=0;
int minn = INF;
int flag = 0;
for (int i=0;i<n;i++)
{
ans += dis(p, v1[i]);
for (int j=0;j<m;j++)
g[i][j] = -dis(v1[i], v2[j]);
}
for (int i = 0; i<n; i++)
{
for (int j = m; j<m+n-1; j++)
g[i][j] = -dis(v1[i], p);
}
nx=ny=m+n-1;
int mc = -KM();
printf("%d\n", mc+ans);
return 0;
}
题解:由于快递员到了酒店之后所有的快递员其实都是一样的了,所以我们建立n-1个虚拟点表示必须有一个快递员进入酒店然后可以继续不断地运送,之后建立n*(n+m-1)条运送路线,设立负边权,然后跑KM得到最小匹配权值。
AC代码:
struct node
{
int x, y;
node(){}
node(int xx,int yy)
{
this->x = xx;
this->y = yy;
}
};
int dis(node x, node y)
{
return abs(x.x - y.x) + abs(x.y - y.y);
}
int main()
{
int n, m,x,y;
scanf("%d%d", &n, &m);
memset(g, 0, sizeof(g));
vector<node>v1,v2;
for (int i=0;i<n;i++){
scanf("%d%d",&x,&y);
v1.push_back(node(x,y));
}
for (int i=0;i<m;i++){
scanf("%d%d",&x,&y);
v2.push_back(node(x,y));
}
node p;
scanf("%d%d",&p.x,&p.y);
int ans=0;
int minn = INF;
int flag = 0;
for (int i=0;i<n;i++)
{
ans += dis(p, v1[i]);
for (int j=0;j<m;j++)
g[i][j] = -dis(v1[i], v2[j]);
}
for (int i = 0; i<n; i++)
{
for (int j = m; j<m+n-1; j++)
g[i][j] = -dis(v1[i], p);
}
nx=ny=m+n-1;
int mc = -KM();
printf("%d\n", mc+ans);
return 0;
}
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