G_M_网络流A_网络吞吐量

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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，LLF），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
*/

G_M_网络流A_网络吞吐量

标签：cout   str   i+1   const   include   http   吞吐量   def   priority   

原文地址：https://www.cnblogs.com/FormerAutumn/p/10355911.html