[SDOI 2009] 晨跑

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=1877

[算法]

        不难看出，第一问要求的是最大流，第二问求的是最小费用最大流

        注意建图时要将每个点拆成入点和出点，防止经过同一个地点多次

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
#define MAXM 40010
const int inf = 2e9;

struct edge
{
        int to,w,cost,nxt;
} e[MAXM << 1];

int i,n,m,tot,a,b,c,S,T,ans1,ans2;
int pre[MAXN << 1],dist[MAXN << 1],incf[MAXN << 1],head[MAXN << 1];

template <typename T> inline void read(T &x)
{
        int f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar())
        {
                if (c == '-') f = -f;
        }
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}
inline void addedge(int u,int v,int w,int cost)
{
        tot++;
        e[tot] = (edge){v,w,cost,head[u]};
        head[u] = tot;
        tot++;
        e[tot] = (edge){u,0,-cost,head[v]};
        head[v] = tot;
}

inline bool spfa()
{
        int i,l,r,u,v,w,cost;
        static int q[MAXN << 1];
        static bool inq[MAXN << 1];
        for (i = 1; i <= 2 * n; i++)
        {
                dist[i] = inf;
                incf[i] = inf;
                inq[i] = false;
        }
        q[l = r = 1] = S;
        inq[S] = true;
        pre[S] = 0;
        dist[S] = 0;
        while (l <= r)
        {
                u = q[l];
                l++;
                inq[u] = false;
                for (i = head[u]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        cost = e[i].cost;
                        if (w && dist[u] + cost < dist[v]) 
                        {
                                dist[v] = dist[u] + cost;
                                incf[v] = min(incf[u],w);
                                pre[v] = i;
                                if (!inq[v])
                                {
                                        inq[v] = true;
                                        q[++r] = v;
                                }
                        }
                }    
        }        
        if (dist[T] != inf) return true;
        else return false;
}
inline void update()
{
        int pos,x = T;
        while (x != S)
        {
                pos = pre[x];
                e[pos].w -= incf[T];
                e[pos ^ 1].w += incf[T];
                x = e[pos ^ 1].to;
        }
        ans1 += incf[T];
        ans2 += dist[T] * incf[T];
}

int main() 
{
        
        read(n); read(m);
        tot = 1;
        addedge(1,n + 1,inf,0);
        addedge(n,2 * n,inf,0);
        for (i = 2; i < n; i++) addedge(i,i + n,1,0);
        for (i = 1; i <= m; i++)
        {
                read(a); read(b); read(c);
                addedge(a + n,b,1,c);        
        }
        S = 1; T = 2 * n;        
        while (spfa()) update();
        printf("%d %d\n",ans1,ans2);
        
        return 0;
    
}