[BZOJ1163][BZOJ1339][Baltic2008]Mafia
试题描述
匪徒准备从一个车站转移毒品到另一个车站,警方准备进行布控. 对于每个车站进行布控都需要一定的代价,现在警方希望使用最小的代价控制一些车站,使得去掉这些车站后,匪徒无法从原定的初始点到达目标点
输入
第一行输入N,M代表车站的总个数,及有多少条双向边连接它们. 2<=n<=200 , 1 <=m<=20000. 第二行给出两个数a,b,代表匪徒的出发点及目标点.1<=a,b<=N,a<>b. 再下来有N行,给出对第i个车站进行布控所需要的Money,其不超过10 000 000 再下来M行,用于描述图的结构.
输出
最少需要多少Money
输入示例
5 6 5 3 2 4 8 3 10 1 5 1 2 2 4 4 5 2 3 3 4
输出示例
5
数据规模及约定
见“输入”
题解
一个点拆成一个“入点”A 和“出点”B,A → B 连一条容量为权值的有向边,无向边拆分成两个有向边,跑一边最小割。
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack> #include <vector> #include <queue> #include <cstring> #include <string> #include <map> #include <set> using namespace std; const int BufferSize = 1 << 16; char buffer[BufferSize], *Head, *tail; inline char Getchar() { if(Head == tail) { int l = fread(buffer, 1, BufferSize, stdin); tail = (Head = buffer) + l; } return *Head++; } int read() { int x = 0, f = 1; char c = Getchar(); while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); } while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); } return x * f; } #define maxn 410 #define maxm 80410 #define oo 2147483647 struct Edge { int from, to, flow; } ; struct Dinic { int n, m, s, t, head[maxn], next[maxm]; Edge es[maxm]; int vis[maxn], Q[maxn], hd, tl; int cur[maxn]; void init(int nn) { n = nn; m = 0; memset(head, -1, sizeof(head)); return ; } void AddEdge(int a, int b, int c) { es[m] = (Edge){ a, b, c }; next[m] = head[a]; head[a] = m++; es[m] = (Edge){ b, a, 0 }; next[m] = head[b]; head[b] = m++; return ; } bool BFS() { memset(vis, 0, sizeof(vis)); vis[s] = 1; hd = tl = 0; Q[++tl] = s; while(hd < tl) { int u = Q[++hd]; for(int i = head[u]; i != -1; i = next[i]) { Edge& e = es[i]; if(e.flow && !vis[e.to]) { vis[e.to] = vis[u] + 1; Q[++tl] = e.to; } } } return vis[t] > 0; } int DFS(int u, int a) { if(u == t || !a) return a; int flow = 0, f; for(int& i = cur[u]; i != -1; i = next[i]) { Edge& e = es[i]; if(vis[e.to] == vis[u] + 1 && (f = DFS(e.to, min(a, e.flow)))) { flow += f; a -= f; e.flow -= f; es[i^1].flow += f; if(!a) return flow; } } return flow; } int MinCut(int ss, int tt) { s = ss; t = tt; int flow = 0; while(BFS()) { for(int i = 1; i <= n; i++) cur[i] = head[i]; flow += DFS(s, oo); } return flow; } } sol; int main() { int n = read(), m = read(); sol.init(n << 1); int s = (read() << 1) - 1, t = read() << 1; for(int i = 1; i <= n; i++) sol.AddEdge((i << 1) - 1, i << 1, read()); for(int i = 1; i <= m; i++) { int u = read(), v = read(); sol.AddEdge(u << 1, (v << 1) - 1, oo); sol.AddEdge(v << 1, (u << 1) - 1, oo); } printf("%d ", sol.MinCut(s, t)); return 0; }