HOJ 2816 Power Line

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Power Line

Time limit:5sec.Submitted:269Memory limit:64MAccepted:84

Source: 2009 China North East Local Contest

Problem Description

While DUT is hosting this NECPC Contest, in order to control the expenditure, careful considerations should be given in many respects, such as the layout of the contest venue. When the contest venue is arranged, each computer requires power input. While the layout of computers and power-points are already fixed, some lines have to be purchased to link the computers with power-points. And, Raven finds out that it is cheaper to buy power lines of the same length than to buy lines of different lengths, so he would buy all the lines with the same length. If Raven wants to save money, would you help him?

Input Details

The first line of the input contains an integer T, which indicates the number of test cases.

For each test case, there are three positive integers in the first row, n, m, p (0 <= n, m <= 200，0 < p < 100). It means that there are n computers and m power-points. The power line costs ￥p per unit.

In the following n rows, i-th line has m positive integers indicating the distances between the i-th computer and all the power points. The distance is less than 10000.

In the last row, there are m positive integers, ci (0 < ci < 5,0 <= i < m), showing No.i power-point can provide power to ci computers in total.

Output Details

For each case, if the layout of computers and power lines are not reasonable, which means you cannot provide power to every computer, just output -1.

Otherwise, output the lowest total price for power lines.

Sample Input

12 2 11 32 21 1

Sample Output

Hint:

You many link the first computer with the first power-points, and link the second computer with the second power-points. This will cost 4(2*2).

二分+多重匹配

水水更健康

#include<cstdio>

#include<cstring>

const int inf=0x7fffffff;

int mat[500][500],link[500][500];

bool map[500][500],usedif[500];

int n,m,p,rmin,rmax;

int lim[500];

bool flag;

bool can(int t)

{

for(int i=1; i<=m; i++)

{

if(usedif[i]==0&&map[t][i])

{

usedif[i]=1;

if(link[i][0]<lim[i])

{

link[i][++link[i][0]]=t;

return true;

}

else

{

for(int j=1; j<=link[i][0]; j++)

if(can(link[i][j]))

{

link[i][j]=t;

return true;

}

return false;

}

bool check(int limit)

{

memset(map,false,sizeof(map));

for(int i=1; i<=m; i++) link[i][0]=0;

for(int i=1; i<=n; i++)

for(int j=1; j<=m; j++)

if(mat[i][j]<=limit) map[i][j]=true;

for(int i=1; i<=n; i++)

{

memset(usedif,0,sizeof(usedif));

if(!can(i)) return false;

}

return true;

}

int find()

{

int high=rmax,low=rmin;

int ans=high;

flag=false;

while(low<high)

{

int mid=(low+high)>>1;

if(check(mid))

{

flag=true;

if(mid<ans) ans=mid;

high=mid;

}

else low=mid+1;

}

return ans;

}

int main()

{

int T;

scanf("%d",&T);

while(T--)

{

scanf("%d%d%d",&n,&m,&p);

rmin=inf;

rmax=-1;

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

{

scanf("%d",&mat[i][j]);

if(mat[i][j]<rmin) rmin=mat[i][j];

if(mat[i][j]>rmax) rmax=mat[i][j];

}

for(int i=1;i<=m;i++) scanf("%d",&lim[i]);

int t=find();

if(flag) printf("%d/n",p*n*t);

else printf("-1/n");

}

return 0;

}

网络流解法，耗时差不多，0.2S左右

#include<cstdio>

#include<cstring>

const int N=1510;

const int M=60001;

const int inf=0x7fffffff;

int head[N];

int lim[N];

int map[N][N];

struct Edge

{

int v,next,w;

} edge[M];

int cnt,s,t;

int n,m,p;

bool flag;

int rmin,rmax,nn;

void addedge(int u,int v,int w)

{

edge[cnt].v=v;

edge[cnt].w=w;

edge[cnt].next=head[u];

head[u]=cnt++;

edge[cnt].v=u;

edge[cnt].w=0;

edge[cnt].next=head[v];

head[v]=cnt++;

}

int sap()

{

int pre[N],cur[N],dis[N],gap[N];

int flow=0,aug=inf,u;

bool flag;

for(int i=0; i<n; i++)

{

cur[i]=head[i];

gap[i]=dis[i]=0;

}

gap[s]=n;

u=pre[s]=s;

while(dis[s]<n)

{

flag=0;

for(int &j=cur[u]; j!=-1; j=edge[j].next)

{

int v=edge[j].v;

if(edge[j].w>0&&dis[u]==dis[v]+1)

{

flag=1;

if(edge[j].w<aug) aug=edge[j].w;

pre[v]=u;

u=v;

if(u==t)

{

flow+=aug;

while(u!=s)

{

u=pre[u];

edge[cur[u]].w-=aug;

edge[cur[u]^1].w+=aug;

}

aug=inf;

}

break;

}

if(flag) continue;

int mindis=n;

for(int j=head[u]; j!=-1; j=edge[j].next)

{

int v=edge[j].v;

if(edge[j].w>0&&dis[v]<mindis)

{

mindis=dis[v];

cur[u]=j;

}

if((--gap[dis[u]])==0)

break;

gap[dis[u]=mindis+1]++;

u=pre[u];

}

return flow;

}

bool check(int limit)

{

cnt=0;

memset(head,-1,sizeof(head));

for(int i=1;i<=nn;i++) addedge(s,i,1);

for(int i=1;i<=m;i++) addedge(i+nn,t,lim[i]);

for(int i=1;i<=nn;i++)

for(int j=1;j<=m;j++)

if(map[i][j]<=limit) addedge(i,j+nn,1);

if(sap()==nn) return true;

return false;

}

int solve()

{

flag=false;

int head=rmin,end=rmax;

int ans=end;

while(head<end)

{

int mid=(head+end)>>1;

if(check(mid))

{

if(ans>mid) ans=mid;

flag=true;

end=mid;

}

else head=mid+1;

}

return ans;

}

int main()

{

int T;

scanf("%d",&T);

while(T--)

{

scanf("%d%d%d",&n,&m,&p);

s=0;

t=n+m+1;

rmin=inf;

rmax=-1;

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

{

scanf("%d",&map[i][j]);

if(map[i][j]<rmin) rmin=map[i][j];

if(map[i][j]>rmax) rmax=map[i][j];

}

for(int i=1;i<=m;i++) scanf("%d",&lim[i]);

nn=n;

n=n+m+2;

int tt=solve();

if(flag) printf("%d/n",p*nn*tt);

else printf("-1/n");

}

return 0;

}