卧薪尝胆,厚积薄发。
CQOI2015 网络吞吐量
Date: Wed Nov 14 21:48:23 CST 2018 In Category: NoCategory

Description:

一个图每个点有容量,要求所有数据沿最短路传送,问同时最多能发送多少数据。
$1\leqslant n\leqslant 500,1\leqslant m\leqslant 10^5$

Solution:

看数据范围像网络流,先把最短路图建出来,然后拆点,跑最大流即可。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cctype>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 510
#define MAXM 100010
typedef long long ll;
struct dijkstra
{
struct edge
{
int to,nxt;
ll v;
}e[MAXM << 1];
int edgenum;
int lin[MAXN];
dijkstra(){edgenum = 0;memset(lin,0,sizeof(lin));}
void add(int a,int b,ll c)
{
e[++edgenum] = (edge){b,lin[a],c};lin[a] = edgenum;
e[++edgenum] = (edge){a,lin[b],c};lin[b] = edgenum;
return;
}
ll d[MAXN];
bool v[MAXN];
int s;
void dij()
{
priority_queue< pair<ll,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;
}
}g,rg;
#define INF 0x3f3f3f3f3f3f3f3f
struct dinic
{
struct edge
{
int to,nxt;
ll f;
}e[(MAXM + MAXN) << 1];
int edgenum;
int lin[MAXN << 1];
dinic(){edgenum = 0;memset(lin,-1,sizeof(lin));}
void add(int a,int b,ll f)
{
e[edgenum] = (edge){b,lin[a],f};lin[a] = edgenum++;
e[edgenum] = (edge){a,lin[b],0};lin[b] = edgenum++;
return;
}
int ch[MAXN << 1];
int s,t;
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);
}
ll flow(int k,int f)
{
if(k == t)return f;
ll 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)
{
ll 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;
}
ll mf()
{
ll ans = 0,r;
while(BFS())while(r = flow(s,INF))ans += r;
return ans;
}
}s;
int main()
{
scanf("%d%d",&n,&m);
int a,b,c;
for(int i = 1;i <= m;++i)
{
scanf("%d%d%d",&a,&b,&c);
g.add(a,b,(ll)c);
rg.add(b,a,(ll)c);
}
g.s = 1;g.dij();
rg.s = n;rg.dij();
s.s = 3;s.t = n << 1;
for(int k = 1;k <= n;++k)
{
int x;
scanf("%d",&x);
s.add(k << 1,k << 1 | 1,(ll)x);
for(int i = g.lin[k];i != 0;i = g.e[i].nxt)
{
if(g.d[k] + g.e[i].v + rg.d[g.e[i].to] == g.d[n])
{
s.add(k << 1 | 1,g.e[i].to << 1,INF);
}
}
}
cout << s.mf() << endl;
return 0;
}
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