卧薪尝胆,厚积薄发。
AHOI2008 上学路线
Date: Fri Oct 12 17:02:05 CST 2018 In Category: NoCategory

Description:

每条边有经过时间和重要度,求从 $1$ 到 $n$ 最短路和删掉重要度之和最小的边使得 $1$ 到 $n$ 最短路变长。
$1\leqslant n\leqslant 500,1\leqslant m\leqslant 124750$

Solution:

看到这么小的数据范围就应该想到网络流,求出最短路图后跑最小割就行了。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 510
#define MAXM 124800
#define INF 0x3f3f3f3f
struct dinic
{
int s,t;
struct edge
{
int to,nxt,f;
}e[MAXM << 1];
int edgenum;
int lin[MAXN];
dinic(){edgenum = 0;memset(lin,-1,sizeof(lin));}
void add(int a,int b,int f)
{
e[edgenum].to = b;e[edgenum].f = f;e[edgenum].nxt = lin[a];lin[a] = edgenum;++edgenum;
e[edgenum].to = a;e[edgenum].f = 0;e[edgenum].nxt = lin[b];lin[b] = edgenum;++edgenum;
return;
}
int ch[MAXN];
bool BFS()
{
queue<int> q;q.push(s);
memset(ch,-1,sizeof(ch));ch[s] = 0;
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return ch[t] != -1;
}
int flow(int k,int f)
{
if(k == t)return f;
int r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f)
{
int l = flow(e[i].to,min(e[i].f,f - r));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r == 0)ch[k] = -1;
return r;
}
int calc()
{
int ans = 0,r;
while(BFS())while(r = flow(s,INF))ans += r;
return ans;
}
}f;
struct dijkstra
{
struct edge
{
int to,nxt,v,f;
}e[MAXM << 1];
int edgenum;
int lin[MAXN];
dijkstra()
{
memset(lin,0,sizeof(lin));
edgenum = 0;
}
void add(int a,int b,int v,int f)
{
++edgenum;e[edgenum].to = b;e[edgenum].v = v;e[edgenum].f = f;e[edgenum].nxt = lin[a];lin[a] = edgenum;
++edgenum;e[edgenum].to = a;e[edgenum].v = v;e[edgenum].f = f;e[edgenum].nxt = lin[b];lin[b] = edgenum;
return;
}
int dis[2][MAXN];
bool v[MAXN];
#define d dis[type]
void calc(int type,int s)
{
priority_queue< pair<int,int> > q;q.push(make_pair(0,s));
memset(d,0x3f,sizeof(d));d[s] = 0;
memset(v,false,sizeof(v));
while(!q.empty())
{
int k = q.top().second;q.pop();
if(v[k])continue;
v[k] = true;
for(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));
}
}
}
return;
}
void work()
{
calc(0,1);calc(1,n);
for(int k = 1;k <= n;++k)
{
for(int i = lin[k];i != 0;i = e[i].nxt)
{
if(dis[0][k] + e[i].v + dis[1][e[i].to] == dis[0][n])
{
f.add(k,e[i].to,e[i].f);
}
}
}
return;
}
}p;
int main()
{
scanf("%d%d",&n,&m);
int a,b,x,y;
for(int i = 1;i <= m;++i)
{
scanf("%d%d%d%d",&a,&b,&x,&y);
p.add(a,b,x,y);
}
p.work();f.s = 1;f.t = n;
cout << p.dis[0][n] << endl;
cout << f.calc() << endl;
return 0;
}
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