bzoj1877 晨跑 费用流

Description

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给出n个点m条边，求不重复地从1出发走到n点最多走多少次，最短走多长的路

Solution

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对于第一问就是拆点的最大流。第二问显然不能单纯用最大流解决了，于是我们每条边引入一个费用的概念，表示单位流量的价格。

连边的时候反向弧的费用要为相反数，那么就是每次找增广路的时候同时找一条费用最小的。因为有负权边所以只能spfa实现

调了一下午我果然还是太弱啊

Code

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#include 
#include 
#include 
#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)
#define erg(i, st) for (int i = ls[st]; i; i = e[i].next)
#define fill(x, t) memset(x, t, sizeof(x))
#define INF 0x3f3f3f3f
#define N 501
#define E 40001
struct edge{int x, y, w, c, next;}e[E];
struct data{int x, t;};
int rc[N][N], ls[N];
inline void addEdge(int &cnt, int x, int y, int w, int c){
    cnt += 1; e[cnt] = (edge){x, y, w, c, ls[x]}; ls[x] = cnt;
    cnt += 1; e[cnt] = (edge){y, x, 0, -c, ls[y]}; ls[y] = cnt;
}
int inQueue[N], dis[N], pre[N], q[N];
inline int spfa(int st, int ed){
    fill(dis, 63);
    fill(inQueue, 0);
    dis[st] = 0;
    inQueue[st] = 1;
    int head = 0, tail = 1;
    q[tail] = st;
    while (head != tail){
        head += 1;
        if (head == N){
            head = 1;
        }
        int now = q[head];
        erg(i, now){
            if (e[i].w > 0 && dis[now] + e[i].c < dis[e[i].y]){
                dis[e[i].y] = dis[now] + e[i].c;
                pre[e[i].y] = i;
                if (!inQueue[e[i].y]){
                    tail += 1;
                    if (tail == N){
                        tail = 1;
                    }
                    q[tail] = e[i].y;
                    inQueue[e[i].y] = 1;
                }
            }
        }
        inQueue[now] = 0;
    }
    return dis[ed] != INF;
}
inline int min(int x, int y){
    return xint ans1 = 0, ans2 = 0;
inline void mcf(int st, int ed){
    int mn = INF;
    for (int i = ed; pre[i]; i = e[pre[i]].x){
        mn = min(mn, e[pre[i]].w);
    }
    for (int i = ed; i; i = e[pre[i]].x){
        e[pre[i]].w -= mn;
        e[pre[i] ^ 1].w += mn;
        ans2 += mn * e[pre[i]].c;
    }
    ans1 += 1;
}
int main(void){
    int n, m;
    scanf("%d%d", &n, &m);
    int st = 1, ed = n + n;
    int edgeCnt = 1;
    fill(ls, 0);
    rep(i, 1, m){
        int x, y, w;
        scanf("%d%d%d", &x, &y, &w);
        addEdge(edgeCnt, x + n, y, 1, w);
    }
    rep(i, 2, n - 1){
        addEdge(edgeCnt, i, i + n, 1, 0);
    }
    addEdge(edgeCnt, st, st + n, INF, 0);
    addEdge(edgeCnt, n, ed, INF, 0);
    while (spfa(st, ed)){
        mcf(st, ed);
    }
    printf("%d %d\n", ans1, ans2);
    return 0;
}