bzoj 1880: [Sdoi2009]Elaxia的路线 最短路

题意

分析

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#define N 1505
#define inf 0x3f3f3f3f
using namespace std;

int cnt,n,m,last
,dis
,ans
,vis
,x1,x2,y1,y2,mx,inq
;
struct edge{int to,next,len,f;}e[N*N];
queue <int> q;

void addedge(int u,int v,int len)
{
e[++cnt].to=v;e[cnt].len=len;e[cnt].next=last[u];last[u]=cnt;
e[++cnt].to=u;e[cnt].len=len;e[cnt].next=last[v];last[v]=cnt;
}

void spfa(int s)
{
memset(dis,inf,sizeof(dis));
memset(vis,0,sizeof(vis));
dis[s]=0;
vis[s]=1;
q.push(s);
while (!q.empty())
{
int u=q.front();
q.pop();
for (int i=last[u];i;i=e[i].next)
if (dis[u]+e[i].len<dis[e[i].to])
{
dis[e[i].to]=dis[u]+e[i].len;
if (!vis[e[i].to])
{
vis[e[i].to]=1;
q.push(e[i].to);
}
}
vis[u]=0;
}
}

void solve()
{
memset(vis,0,sizeof(vis));
q.push(y1);
vis[y1]=1;
while (!q.empty())
{
int u=q.front();
q.pop();
for (int i=last[u];i;i=e[i].next)
if (dis[e[i].to]+e[i].len==dis[u])
{
e[i].f=e[i^1].f=1;
if (!vis[e[i].to])
{
vis[e[i].to]=1;
q.push(e[i].to);
}
}
}
}

void get_ans()
{
memset(vis,0,sizeof(vis));
q.push(y2);
vis[y2]=1;
inq[y2]=1;
while (!q.empty())
{
int u=q.front();
q.pop();
for (int i=last[u];i;i=e[i].next)
if (dis[e[i].to]+e[i].len==dis[u])
{
if (e[i].f&&ans[u]+e[i].len>ans[e[i].to])
{
ans[e[i].to]=ans[u]+e[i].len;
mx=max(mx,ans[e[i].to]);
if (!inq[e[i].to])
{
inq[e[i].to]=1;
q.push(e[i].to);
}
}
if (!vis[e[i].to])
{
vis[e[i].to]=1;
inq[e[i].to]=1;
q.push(e[i].to);
}
}
inq[u]=0;
}
}

int main()
{
scanf("%d%d",&n,&m);
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
cnt=1;
for (int i=1;i<=m;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
addedge(x,y,z);
}
spfa(x1);
solve();
spfa(x2);
get_ans();
printf("%d",mx);
return 0;
}