卧薪尝胆,厚积薄发。
JLOI2011 飞行路线
Date: Wed Aug 01 14:13:57 CST 2018 In Category: NoCategory

Description:

一个图,可以将不超过 $k$ 条边边权改成 $0$ ,问从 $s$ 到 $t$ 的最短路。
$1\le n \le 10^4$ $1\le m \le 5\times 10^4$ $1\le k \le 10$

Solution:

分层图。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
int n,m,k;
int s,t;
#define MAXM 50010
#define MAXN 10010
#define MAXK 15
struct edge
{
int to,nxt,v;
}e[(MAXM << 1) * MAXK * 2];
int edgenum = 0;
int lin[MAXN * MAXK] = {0};
inline void add(int a,int b,int c)
{
++edgenum;e[edgenum].to = b;e[edgenum].v = c;e[edgenum].nxt = lin[a];lin[a] = edgenum;
return;
}
int d[MAXN * MAXK];
bool v[MAXN * MAXK];
inline int read()
{
register int res = 0;
register char c = getchar();
while(!isdigit(c))c = getchar();
while(isdigit(c))
{
res = (res << 1) + (res << 3) + c - '0';
c = getchar();
}
return res;
}
int main()
{
scanf("%d%d%d",&n,&m,&k);
s = read();t = read();
++s;++t;
register int a,b,c;
for(register int i = 1;i <= m;++i)
{
a = read();b = read();c = read();
++a;++b;
for(register int j = 0;j <= k;++j)
{
add(a + j * n,b + j * n,c);
add(b + j * n,a + j * n,c);
}
for(register int j = 0;j < k;++j)
{
add(a + j * n,b + (j + 1) * n,0);
add(b + j * n,a + (j + 1) * n,0);
}
}
priority_queue<pair<int,int> > q;
memset(d,0x3f,sizeof(d));
d[s] = 0;
q.push(make_pair(0,s));
while(!q.empty())
{
register int k = q.top().second;q.pop();
if(v[k])continue;
v[k] = true;
for(register int i = lin[k];i != 0;i = e[i].nxt)
{
if(d[e[i].to] > d[k] + e[i].v)
{
d[e[i].to] = d[k] + e[i].v;
q.push(make_pair(-d[e[i].to],e[i].to));
}
}
}
int ans = 0x3f3f3f3f;
for(register int i = 0;i <= k;++i)
{
ans = min(ans,d[t + i * n]);
}
printf("%d\n",ans);
return 0;
}
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