G_M_网络流A_网络吞吐量
调了两天的代码，到最后绝望地把I64d改成lld就过了，我真的是醉了。
网络吞吐量
题面：给出一张（n个点，m条边）带权（点权边权均有）无向图，点权为每个点每秒可以接受发送的最大值，边权为花费，保证数据沿着最短路径从1发送到n。
题解：因为保证数据沿着最短路发送，所以可以求出1到n的最短路，然后将符合最短路的边全部加到网络图（一定是符合条件的所有边，而非最短路上的边，因为最短路长度唯一，而路径不唯一，故点权不唯一，所以应该将所有符合条件的边都加入），这些边的容量为INF，因为流量限制在点上，所以最短路边权不限制流量，限制工作交给点来完成。将一个点拆成入点和出点，入点到出点的边权为点权以达到限流作用，最后跑一次最大流即为所求。
PS：这题是CQOI2015的题，不得不说这题作为专题的A是有道理的，因为是最短路+网络流拆点的模板，比较和谐。~鬼知道I64d会WA？？？~除此，还了解到了SPFA的两个小优化（SLF，LLL），可以说收获不小。~ISAP是真滴好写~
#pragma comment(linkerr, "/STACK: 1024000000,1024000000")
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define em emplace
#define pii pair<int,int>
#define de(x) cout << #x << " = " << x << endl
#define clr(a,b) memset(a,b,sizeof(a))
#define INF (0x3f3f3f3f)
#define LINF ((long long)(0x3f3f3f3f3f3f3f3f))
#define F first
#define S second
using namespace std;
const int N = 5006;
const int M = 1e5 + 15;
typedef long long ll;
typedef pair<ll, int> pli;
struct Edge
{
int u, v, nxt;
ll w;
};
Edge e1[M<<2], e2[M<<2];
int h1[N<<2], h2[N<<2], ect1, ect2;
ll dis[N<<2];
int d[N<<2], gap[N<<2], cur[N<<2];
bool vis[N<<2];
int n, m, ct;
int a[M], b[M];
ll c[M];
ll val[N];
void init()
{
clr(vis,0);
clr(h1,-1); clr(h2,-1);
ect1 = ect2 = 0;
}
void _add1( int u, int v, ll w )
{
e1[ect1].u = u;
e1[ect1].v = v;
e1[ect1].nxt = h1[u];
e1[ect1].w = w;
h1[u] = ect1++;
}
void _add2( int u, int v, ll w )
{
e2[ect2].u = u;
e2[ect2].v = v;
e2[ect2].nxt = h2[u];
e2[ect2].w = w;
h2[u] = ect2++;
}
void dij( int s, int t )
{
for ( int i = 1; i <= n; i ++ )
dis[i] = LINF;
dis[t] = 0;
priority_queue<pli> q;
q.push( {0ll,t} );
while ( !q.empty() )
{
pli nw = q.top(); q.pop();
int u = nw.S;
if ( vis[u] ) continue;
vis[u] = 1;
for ( int i = h1[u]; i+1; i = e1[i].nxt )
{
int v = e1[i].v;
if ( dis[v] > dis[u] + e1[i].w )
{
dis[v] = dis[u] + e1[i].w;
q.push( {-dis[v], v} );
}
}
}
}
void bfs( int s, int t )
{
queue<int> q;
clr(d,0); clr(gap,0);
++gap[d[t] = 1];
ct = 1;
q.push(t);
while ( !q.empty() )
{
int u = q.front(); q.pop();
for ( int i = h2[u]; i+1; i = e2[i].nxt )
{
int v = e2[i].v;
if ( d[v] ) continue;
++gap[d[v] = d[u]+1];
q.push(v);
ct ++;
}
}
}
ll aug( int u, int s, int t, ll mi )
{
if ( u == t ) return mi;
ll flw = 0;
for ( int &i = cur[u]; i+1; i = e2[i].nxt )
{
int v = e2[i].v;
if ( d[u] == d[v] + 1 )
{
ll tp = aug( v, s, t, min(mi,e2[i].w) );
flw += tp, mi -= tp, e2[i].w -= tp, e2[i^1].w += tp;
if ( !mi ) return flw;
}
}
if ( !(--gap[d[u]]) ) d[s] = ct+1;
++gap[++d[u]], cur[u] = h2[u];
return flw;
}
ll mxflw( int s, int t )
{
bfs(s,t);
for ( int i = 1; i <= 2*n; i ++ )
cur[i] = h2[i];
ll res = aug( s, s, t, LINF );
while ( d[s] <= ct )
res += aug( s, s, t, LINF );
return res;
}
void build( int s, int t )
{
queue<int> q;
q.push(s);
clr(vis,0);
vis[s] = 1;
while ( !q.empty() )
{
int u = q.front(); q.pop();
for ( int i = h1[u]; i+1; i = e1[i].nxt )
{
int v = e1[i].v;
if ( dis[u] == dis[v] + e1[i].w )
{
_add2( u+n, v, LINF ), _add2( v, u+n, 0ll );
if ( !vis[v] ) q.push(v), vis[v] = 1;
}
}
}
}
int main()
{
init();
scanf("%d%d", &n, &m);
for ( int i = 0; i < m; i ++ )
{
scanf("%d%d%lld", &a[i], &b[i], &c[i]);
_add1(a[i],b[i],c[i]); _add1(b[i],a[i],c[i]);
}
dij(1,n);
build(1,n);
for ( int i = 1; i <= n; i ++ )
scanf("%lld", &val[i]);
for ( int i = 2; i < n; i ++ )
{
_add2( i, i+n, val[i]);
_add2( i+n, i, 0ll );
}
int S = 1, T = n;
_add2( 1, 1+n, LINF ); _add2( 1+n, 1, 0ll );
ll mx = mxflw( S, T );
printf("%lld\n", mx);
return 0;
}
/*
3 2
1 2 1
2 3 1
1
2
3
*/
