Topic Portal (internal title 149)
Input Format
Each test point of the first behavior four positive integer $ n, b, s, m $, as meaning the subject.
Next $ m $ rows of three non-negative integer $ u, v, l $, $ represents from point to point U $ $ $ V L has a weight value of $ $ directed edge, the data is guaranteed FIG. strongly connected, i.e. can walk between any two points each.
Output Format
A data output line of each non-negative integer answer.
Sample
Sample input 1:
5 4 2 10
5 2 1
2 5 1
3 5 5
4 5 0
1 5 1
2 3 1
3 2 5
2 4 5
2 1 1
3 4 2
Sample output 1:
13
Sample input 2:
5 4 2 10
5 2 1
2 5 1
3 5 5
4 5 10
1 5 1
2 3 1
3 2 5
2 4 5
2 1 1
3 4 2
Sample Output 2:
24
Data range and tips
$ $ 5 per test points, each test point data range is as follows:
In addition, a number may be a uniform value of $ s $ relatively small data point appears.
For all test points are $ m \ leqslant 50,000,1 \ leqslant s \ leqslant b <n, 0 \ leqslant l \ leqslant 10,000 $, given strong and legitimate communication to FIG.
answer
Positive and negative first find the shortest path to the headquarters of each point, you can build the reverse side.
Of the formula can be found at every point of the positive and negative price for its shortest length and multiplying it has several divisions at $ -1 $ subprojects.
利用贪心的思想，小的放在一起，大的放在一起一定更优，于是可以排个序。
接着考虑$DP$，设$dp[i][j]$表示选到$i$，分了$j$组的最小代价，但是发现时间复杂度是$\Theta(n^3)$的，接着考虑优化。
在最优解中$dis$从小到大依次划分得到的段的长度一定是单调不增的，所以只有$[i-\frac{i}{j},i)$才能更新$dp[i][j]$。
那么现在算法的时间复杂度就是：$\Theta(n^2(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}))=\Theta(n^2\log n)$。
由于时限为$3s$所以跑过去绰绰有余。
时间复杂度：$\Theta(n^2\log n)$。
期望得分：$100$分。
实际得分：$100$分。
代码时刻
#include<bits/stdc++.h>
using namespace std;
struct rec{int nxt,to,w;}e[100001];
int head[2][5001],cnt;
int n,b,s,m;
int dis[2][5001];
bool vis[5001];
long long sum[5001],dp[5001][5001];
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>q;
void add(bool id,int x,int y,int w)
{
e[++cnt].nxt=head[id][x];
e[cnt].to=y;
e[cnt].w=w;
head[id][x]=cnt;
}
void Dij0()
{
q.push(make_pair(0,b+1));
dis[0][b+1]=0;
while(q.size())
{
int x=q.top().second;q.pop();
if(vis[x])continue;vis[x]=1;
for(int i=head[0][x];i;i=e[i].nxt)
if(dis[0][e[i].to]>dis[0][x]+e[i].w)
{
dis[0][e[i].to]=dis[0][x]+e[i].w;
q.push(make_pair(dis[0][e[i].to],e[i].to));
}
}
}
void Dij1()
{
memset(vis,0,sizeof(vis));
q.push(make_pair(0,b+1));
dis[1][b+1]=0;
while(q.size())
{
int x=q.top().second;q.pop();
if(vis[x])continue;vis[x]=1;
for(int i=head[1][x];i;i=e[i].nxt)
if(dis[1][e[i].to]>dis[1][x]+e[i].w)
{
dis[1][e[i].to]=dis[1][x]+e[i].w;
q.push(make_pair(dis[1][e[i].to],e[i].to));
}
}
}
int main()
{
scanf("%d%d%d%d",&n,&b,&s,&m);
for(int i=1;i<=m;i++)
{
int a,b,l;
scanf("%d%d%d",&a,&b,&l);
add(0,a,b,l);add(1,b,a,l);
}
memset(dis,0x3f,sizeof(dis));
Dij0();Dij1();
for(int i=1;i<=b;i++)
sum[i]=dis[0][i]+dis[1][i];
sort(sum+1,sum+b+1);
for(int i=1;i<=b;i++)sum[i]+=sum[i-1];
memset(dp,0x3f,sizeof(dp));
dp[0][0]=0;
for(int i=1;i<=b;i++)
for(int j=1;j<=s;j++)
for(int k=i-i/j;k<i;k++)
{
if((sum[i]-sum[k])*(i-k-1)>=dp[i][j])break;
dp[i][j]=min(dp[i][j],dp[k][j-1]+(sum[i]-sum[k])*(i-k-1));
}
printf("%lld",dp[b][s]);
return 0;
}
rp++