P2153 晨跑,費用流裸題

晨跑

題目連接

https://www.luogu.org/problemnew/show/P2153

題解

求最大不相交路徑數,並在路徑數最大前提下,求總路程最短.

太裸了.

求不相交路徑數:將除$1,n$兩點外的所有點拆分,中間連一條容量爲$1$,費用爲$0$的邊.然後所有的原邊$u \rightarrow v$視作從$u$的出點連向$v$的入點的一條費用爲路程,容量爲$1$的邊.

從$1 \rightarrow n$跑最小費用最大流即是答案.

代碼

// luogu-judger-enable-o2
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)

const int inf = 0x3f3f3f3f;
const int mm = 111111;
const int maxn = 999;
int node,src,dest,edge;
int ver[mm],flow[mm],cst[mm],nxt[mm];
int head[maxn],work[maxn],dis[maxn],q[maxn];
int tot_cost;
void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<node; ++i)head[i]=-1;
    edge=0;
    tot_cost = 0;
}
void add_edge(int u,int v,int c,int cost)
{
    ver[edge]=v,flow[edge]=c,nxt[edge]=head[u],cst[edge]=cost,head[u]=edge++;
    ver[edge]=u,flow[edge]=0,nxt[edge]=head[v],cst[edge]=-cost,head[v]=edge++;
}
int ins[maxn];
int pre[maxn];
bool Dinic_spfa()
{
    memset(ins,0,sizeof(ins));
    memset(dis,inf,sizeof(dis));
    memset(pre,-1,sizeof(pre));
    std::queue<int> Q;
    Q.push(src);
    dis[src] = 0,ins[src] = 1;
    pre[src] = -1;
    while(!Q.empty()){
        int u = Q.front();Q.pop();
        ins[u] = 0;
        for(int e = head[u];e != -1;e = nxt[e]){
            int v = ver[e];
            if(!flow[e]) continue;
            if(dis[v] > dis[u] + cst[e]){
                dis[v] = dis[u] + cst[e];
                pre[v] = e;
                if(!ins[v]) ins[v] = 1,Q.push(v);
            }
        }
    }
    return dis[dest] < inf;
}
int Dinic_flow()
{
    int i,ret=0,delta=inf;
    while(Dinic_spfa())
    {
        for(int i=pre[dest];i != -1;i = pre[ver[i^1]])
    		delta = std::min(delta,flow[i]);
        for(int i=pre[dest];i != -1;i = pre[ver[i^1]])
    		flow[i] -= delta,flow[i^1] += delta;
        ret+=delta;
        tot_cost += dis[dest]*delta;
    }
    return ret;
}
int n,m;
int main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n >> m;
    prepare(2*n,0,2*n-1);
    for(int i = 1;i <= n;++i) {
        if(i == 1 || i == n)
            add_edge(i-1,i-1+n,inf,0);
        else 
            add_edge(i-1,i-1+n,1,0);
    }
    for(int i = 1;i <= m;++i) {
        int a,b,c;
        std::cin >> a >> b >> c;
        add_edge(a-1+n,b-1,1,c);
    }
    int myflow = Dinic_flow();
    std::cout << myflow << " " << tot_cost << std::endl;
    return 0;
}