title
BZOJ 2324
LUOGU 4542
Description
Rockets Pikachu was snatched by the evil plot! The three bad guys gave little wisdom left a red fruit provocation! In order to Pikachu, but also for justice, Ash and his friends were duty-bound to set foot on the road to rescue Pikachu.
There exists a strip M two-way road between the N stronghold stronghold Rockets. From 1 to N positions, respectively, numerals. Ash and his party really new K people from the town, to rescue trapped Pikachu N number of strongholds. For convenience, we will really new town regarded as strongholds 0, K individuals beginning at 0 points.
Due to heavy deployment Rockets, in order to destroy the stronghold K number, you must first destroy the order of K-1 No. 1 to the base, and, if the K-1 No. stronghold not destroyed, either due to the defensive nature of the chain, Ash and his party people entering the stronghold K, will be found, and have serious consequences. Therefore, before K-1 No. stronghold was destroyed, people are not able to go through any number K stronghold.
For simplicity, we ignore the part of the battle, Ash and his party anyone through K K No. No. stronghold that is considered strongholds were destroyed. The stronghold was destroyed can still be passed.
K individual is split up, as long as there is any one person after the K-1 No. stronghold was destroyed, after K No. stronghold, K No. stronghold was destroyed. Obviously, as long as No. N stronghold was destroyed, Pikachu would be saved.
Wild road is not safe, so little wisdom and his party at the same time want to destroy strongholds number N rescued Pikachu minimize the total length of at least K individuals through road.
Please help you design a little wisdom the best rescue plan now!
Input
The first line contains three positive integers N, M, K. It represents a total of N + 1 positions, respectively, undirected edges numbered from 0 to N, and the M. Ash and his party were beginning K individuals are located 0:00.
Next M rows, each row of three non-negative integer, the integer i-th row of Ai, Bi, Li. It indicates the presence of a base numbers from Ai to Bi length of a road base number of Li.
Output
It contains only one integer S, to rescue the smallest sum pikaqiu road that needs to elapse.
Sample Input
3 4 2
0 1 1
1 2 1
2 3 100
0 3 1
Sample Output
3
[] sample shows
a small intellectual and Xiaoxia together to rescue Pikachu. In the best scenario, starting with the small Chilean town went really new No. 1 o'clock, then went Depot II. When Ash succeeded in destroying the stronghold No. 2, Xiaoxia departure directly to the base 3, rescue Pikachu from a really new town.
HINT
To 100% data satisfy N ≤ 150, M ≤ 20 000, 1 ≤ K ≤ 10, Li ≤ 10 000, a line must be able to ensure Ash Pikachu rescued. As for why K ≤ 10, you may think that eventually the call of the ash, the ash, Xiaoxia, Xiaogang, Xiaojian, a small remote, a small victory, little light, Alice, Tin Tong, as well as travel to Japan the black sergeant, went to war with the Rockets.
Source
analysis
Simplify the meaning of the questions:
After using up \ (K \) section through \ (0 \) path to cover all the points.
Pretreatment, provided \ (d [i] [j ] \) represents the \ (I \) to \ (J \) without greater than \ (\ max (i, j ) \) shortest point, clearly you can use \ (floyed \) requirements.
Built map:
- Each point split point, even edge;
- Source to the point of even flow 0 \ (k \) side of the cost of 0, indicates that up after 0:00 \ (k \) times;
- A flow source connected to the remaining side of point costs for each point of 0;
- Each \ (I \) is the point \ (j (j> i) \) connected charges flow 1 \ (d [i] [j ] \) side;
- Each point of the 1 charges flow connected to the sink edge points 0;
- It can run a minimal cost.
code
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10,maxm=1e6+10,inf=0x3f3f3f3f;
char buf[1<<15],*fs,*ft;
inline char getc() { return (ft==fs&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),ft==fs))?0:*fs++; }
template<typename T>inline void read(T &x)
{
x=0;
T f=1, ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
template<typename T>inline void write(T x)
{
if (!x) { putchar('0'); return ; }
if (x<0) putchar('-'), x=-x;
T num=0, ch[20];
while (x) ch[++num]=x%10+48, x/=10;
while (num) putchar(ch[num--]);
}
int ver[maxm<<1],edge[maxm<<1],Next[maxm<<1],cost[maxm<<1],head[maxn],len=1;
inline void add(int x,int y,int z,int c)
{
ver[++len]=y,edge[len]=z,cost[len]=c,Next[len]=head[x],head[x]=len;
ver[++len]=x,edge[len]=0,cost[len]=-c,Next[len]=head[y],head[y]=len;
}
int s,t;
int dist[maxn],incf[maxn],pre[maxn];
bool vis[maxn];
inline bool spfa()
{
memset(dist,0x3f,sizeof(dist));
memset(vis,0,sizeof(vis));
queue<int>q;q.push(s);
dist[s]=0,vis[s]=1,incf[s]=1<<30;
while (!q.empty())
{
int x=q.front();
q.pop();
vis[x]=0;
for (int i=head[x]; i; i=Next[i])
{
if (!edge[i]) continue;
int y=ver[i];
if (dist[y]>dist[x]+cost[i])
{
dist[y]=dist[x]+cost[i];
incf[y]=min(incf[x],edge[i]);
pre[y]=i;
if (!vis[y]) q.push(y),vis[y]=1;
}
}
}
if (dist[t]==inf) return false;
else return true;
}
long long maxflow,ans;
inline void update()
{
int x=t;
while (x!=s)
{
int i=pre[x];
edge[i]-=incf[t];
edge[i^1]+=incf[t];
x=ver[i^1];
}
maxflow+=incf[t];
ans+=dist[t]*incf[t];
}
int d[201][201];
int main()
{
int n,m,k;read(n);read(m);read(k);
for (int i=0; i<=n; ++i)
for (int j=0; j<=n; ++j)
if (i!=j) d[i][j]=inf;
for (int i=1,x,y,z; i<=m; ++i) read(x),read(y),read(z),d[x][y]=d[y][x]=min(d[x][y],z);
for (int k=0; k<=n; ++k)
for (int i=0; i<=n; ++i)
for (int j=0; j<=n; ++j)//这道题比较sb地把0点搞成图中的一个点
if (k<max(i,j)) d[i][j]=min(d[i][j],d[i][k]+d[k][j]);//floyed，保证经过点最大值<max(i,j)
s=n<<1|1,t=s+1;
for (int i=0; i<=n; ++i)
{
add(s,i,!i?k:1,0),add(i+n,t,1,0);//0点要保证k个人都过去
for (int j=i+1; j<=n; ++j) add(i,j+n,1,d[i][j]);//i->j
}
while (spfa()) update();//好像没有上下界，只有上界乎？
write(ans),puts("");
return 0;
}