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  • [BZOJ1163][BZOJ1339][Baltic2008]Mafia

    [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;
    }
    
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  • 原文地址:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5475630.html
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