bzoj-1834 network 网络扩容 【网络流】

　　这题就是复习下网络流。

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define mp make_pair
#define pb push_back
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
using namespace std;
typedef long long i64;
typedef pair<int, int> pii;
const int inf = ~0U >> 1;
const i64 INF = ~0ULL >> 1;
//********************************

const int maxn = 1005, maxm = 5005;

struct Ed {
    int u, v, nx, c, w; Ed() {}
    Ed(int _u, int _v, int _nx, int _c, int _w) :
        u(_u), v(_v), nx(_nx), c(_c), w(_w) {}
} E[maxm << 2];
int G[maxn], edtot, cur[maxn];
void addedge(int u, int v, int c, int w) {
    E[edtot] = Ed(u, v, G[u], c, w);
    G[u] = edtot++;
    E[edtot] = Ed(v, u, G[v], 0, -w);
    G[v] = edtot++;
}

int level[maxn], s , t;
bool bfs() {
    static int que[maxn]; int qh(0), qt(0);
    clr(level);
    level[que[++qt] = s] = 1;
    while (qh != qt) {
        int x = que[++qh];
        for (int i = G[x]; i != -1; i = E[i].nx) if (!level[E[i].v] && E[i].c) {
            level[que[++qt] = E[i].v] = level[x] + 1;
        }
    }
    return !!level[t];
}
int dfs(int u, int rm) {
    if (u == t) return rm;
    int rm1 = rm;
    for (int &i = cur[u]; i != -1; i = E[i].nx) {
        if (level[E[i].v] == level[u] + 1 && E[i].c) {
            int flow = dfs(E[i].v, min(rm, E[i].c));
            E[i].c -= flow, E[i ^ 1].c += flow;
            if ((rm -= flow) == 0) break;
        }
    }
    if (rm1 == rm) level[u] = 0;
    return rm1 - rm;
}

int dis[maxn], vis[maxn];
bool spfa() {
    static int que[maxn]; int qh(0), qt(0);
    memset(dis, 0x3f, sizeof(dis));
    clr(vis);
    dis[que[++qt] = s] = 0, vis[s] = 1;
    while (qh != qt) {
        int x = que[++qh]; vis[x] = 0;
        for (int i = G[x]; i != -1; i = E[i].nx) {
            if (E[i].c && dis[E[i].v] > dis[x] + E[i].w) {
                dis[E[i].v] = dis[x] + E[i].w;
                if (!vis[E[i].v]) que[++qt] = E[i].v, vis[E[i].v] = 1;
            }
        }
    }
    return dis[t] != 0x3f3f3f3f;
}
int ret;
int dinic(int u, int rm) {
    vis[u] = 1;
    if (u == t) return rm;
    int rm1 = rm;
    for (int i = G[u]; i != -1; i = E[i].nx) {
        if (dis[E[i].v] == dis[u] + E[i].w && E[i].c && !vis[E[i].v]) {
            int flow = dinic(E[i].v, min(rm , E[i].c));
            E[i].c -= flow, E[i ^ 1].c += flow;
            ret += flow * E[i].w;
            if ((rm -= flow) == 0) break;
        }
    }
    if (rm1 == rm) dis[u] = 0;
    return rm1 - rm;
}

int main() {
    int n, m, K;
    scanf("%d%d%d", &n, &m, &K);
    memset(G, -1, sizeof(G));
    rep(i, 1, m) {
        int x, y, c, w;
        scanf("%d%d%d%d", &x, &y, &c, &w);
        addedge(x, y, c, w);
        addedge(x, y, c, 0);
    }
    s = 0, t = n + 1; addedge(s, 1, 0x3f3f3f3f, 0), addedge(n, t, 0x3f3f3f3f, 0);
    int ans(0);
    while (bfs()) memcpy(cur, G, sizeof(G)), ans += dfs(s, 0x3f3f3f3f);
    printf("%d ", ans >> 1);
    for (int i = 0; i <= edtot; i += 2) {
        if (E[i].w == 0) E[i].c += E[i ^ 1].c, E[i ^ 1].c = 0;
        else E[i].c = 0x3f3f3f3f, E[i ^ 1].c = 0;
        if (E[i].u == s) E[i].c = (ans >> 1) + K;
    }
    while (spfa()) clr(vis), dinic(s, 0x3f3f3f3f);
    printf("%d
", ret);
}