2017湖南多校第一场-B(1002): Bones’s Battery

Problem B | limit 5 seconds

Bones’s Battery

Bones is investigating what electric shuttle is appropriate for his mom’s school district vehicle.

Each school has a charging station. It is important that a trip from one school to any other be

completed with no more than K rechargings. The car is initially at zero battery and must always

be recharged at the start of each trip; this counts as one of the K rechargings. There is at most

one road between each pair of schools, and there is at least one path of roads connecting each pair

of schools. Given the layout of these roads and K, compute the necessary range required of the

electric shuttle.

Input

Input begins with a line with one integer T (1  T  50) denoting the number of test cases. Each

test case begins with a line containing three integers N, K, and M (2  N  100, 1  K  100),

where N denotes the number of schools, K denotes the maximum number of rechargings permitted

per trip, and M denotes the number of roads. Next follow M lines each with three integers ui, vi,

and di (0  ui; vi < N, ui 6= vi, 1  di  109) indicating that road i connects schools ui and vi

(0-indexed) bidirectionally with distance di.

Output

For each test case, output one line containing the minimum range required.

Sample Input Sample Output

2

4 2 4

0 1 100

1 2 200

2 3 300

3 0 400

10 2 15

0 1 113

1 2 314

2 3 271

3 4 141

4 0 173

5 7 235

7 9 979

9 6 402

6 8 431

8 5 462

0 5 411

1 6 855

2 7 921

3 8 355

4 9 113

300

688

解题思路：先用floyd跑一遍求得最短路，再二分电量，用dp[i][j]表示i到j需充电的次数，在用floyd对dp跑一遍，找出dp中最大值，跟k进行比较，最后得出答案

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long LL;
#define INF 100000000000L
LL dist[105][105];
LL dp[105][105];
int n,k,m;

void init()
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
dist[i][j]=INF;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
dp[i][j]=INF;
}

void floyd1()
{
for(int i=0;i<n;i++)
dist[i][i]=0;
for(int k=0;k<n;k++)
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
}

void floyd2()
{
for(int i=0;i<n;i++)
dp[i][i]=0;
for(int k=0;k<n;k++)
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{

scanf("%d%d%d",&n,&k,&m);
init();
int u,v,d;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&d);
dist[u][v]=dist[v][u]=d;
}
floyd1();
LL Min=1,Max=INF;
LL ans=Max;
LL Mid;
while(Min<=Max)
{
Mid=(Min+Max)/2;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
dp[i][j]=dist[i][j]<=Mid?1:INF;
floyd2();
LL d=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(dp[i][j]>d) d=dp[i][j];
if(d<=k)
{
ans=Mid;
Max=Mid-1;
}else
{
Min=Mid+1;
}
}
printf("%lld\n",ans);
}
return 0;
}