bzoj-1834 network 网络扩容 【网络流】
这题就是复习下网络流。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define mp make_pair 6 #define pb push_back 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U >> 1; 14 const i64 INF = ~0ULL >> 1; 15 //******************************** 16 17 const int maxn = 1005, maxm = 5005; 18 19 struct Ed { 20 int u, v, nx, c, w; Ed() {} 21 Ed(int _u, int _v, int _nx, int _c, int _w) : 22 u(_u), v(_v), nx(_nx), c(_c), w(_w) {} 23 } E[maxm << 2]; 24 int G[maxn], edtot, cur[maxn]; 25 void addedge(int u, int v, int c, int w) { 26 E[edtot] = Ed(u, v, G[u], c, w); 27 G[u] = edtot++; 28 E[edtot] = Ed(v, u, G[v], 0, -w); 29 G[v] = edtot++; 30 } 31 32 int level[maxn], s , t; 33 bool bfs() { 34 static int que[maxn]; int qh(0), qt(0); 35 clr(level); 36 level[que[++qt] = s] = 1; 37 while (qh != qt) { 38 int x = que[++qh]; 39 for (int i = G[x]; i != -1; i = E[i].nx) if (!level[E[i].v] && E[i].c) { 40 level[que[++qt] = E[i].v] = level[x] + 1; 41 } 42 } 43 return !!level[t]; 44 } 45 int dfs(int u, int rm) { 46 if (u == t) return rm; 47 int rm1 = rm; 48 for (int &i = cur[u]; i != -1; i = E[i].nx) { 49 if (level[E[i].v] == level[u] + 1 && E[i].c) { 50 int flow = dfs(E[i].v, min(rm, E[i].c)); 51 E[i].c -= flow, E[i ^ 1].c += flow; 52 if ((rm -= flow) == 0) break; 53 } 54 } 55 if (rm1 == rm) level[u] = 0; 56 return rm1 - rm; 57 } 58 59 int dis[maxn], vis[maxn]; 60 bool spfa() { 61 static int que[maxn]; int qh(0), qt(0); 62 memset(dis, 0x3f, sizeof(dis)); 63 clr(vis); 64 dis[que[++qt] = s] = 0, vis[s] = 1; 65 while (qh != qt) { 66 int x = que[++qh]; vis[x] = 0; 67 for (int i = G[x]; i != -1; i = E[i].nx) { 68 if (E[i].c && dis[E[i].v] > dis[x] + E[i].w) { 69 dis[E[i].v] = dis[x] + E[i].w; 70 if (!vis[E[i].v]) que[++qt] = E[i].v, vis[E[i].v] = 1; 71 } 72 } 73 } 74 return dis[t] != 0x3f3f3f3f; 75 } 76 int ret; 77 int dinic(int u, int rm) { 78 vis[u] = 1; 79 if (u == t) return rm; 80 int rm1 = rm; 81 for (int i = G[u]; i != -1; i = E[i].nx) { 82 if (dis[E[i].v] == dis[u] + E[i].w && E[i].c && !vis[E[i].v]) { 83 int flow = dinic(E[i].v, min(rm , E[i].c)); 84 E[i].c -= flow, E[i ^ 1].c += flow; 85 ret += flow * E[i].w; 86 if ((rm -= flow) == 0) break; 87 } 88 } 89 if (rm1 == rm) dis[u] = 0; 90 return rm1 - rm; 91 } 92 93 int main() { 94 int n, m, K; 95 scanf("%d%d%d", &n, &m, &K); 96 memset(G, -1, sizeof(G)); 97 rep(i, 1, m) { 98 int x, y, c, w; 99 scanf("%d%d%d%d", &x, &y, &c, &w); 100 addedge(x, y, c, w); 101 addedge(x, y, c, 0); 102 } 103 s = 0, t = n + 1; addedge(s, 1, 0x3f3f3f3f, 0), addedge(n, t, 0x3f3f3f3f, 0); 104 int ans(0); 105 while (bfs()) memcpy(cur, G, sizeof(G)), ans += dfs(s, 0x3f3f3f3f); 106 printf("%d ", ans >> 1); 107 for (int i = 0; i <= edtot; i += 2) { 108 if (E[i].w == 0) E[i].c += E[i ^ 1].c, E[i ^ 1].c = 0; 109 else E[i].c = 0x3f3f3f3f, E[i ^ 1].c = 0; 110 if (E[i].u == s) E[i].c = (ans >> 1) + K; 111 } 112 while (spfa()) clr(vis), dinic(s, 0x3f3f3f3f); 113 printf("%d\n", ret); 114 }
