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  • Poj2749:Building roads

    题意

    有 N 个牛栏,现在通过一条通道(s1,s2)要么连到s1,要么连到s2,把他们连起来,他们之间有一些约束关系,一些牛栏不能连在同一个点,一些牛栏必须连在同一个点,现在问有没有可能把他们都连好,而且满足所有的约束关系,如果可以,输出距离最大的两个牛栏之间距离最小值(两点距离是指哈密顿距离)

    Sol

    二分答案+(2-SAT)判定
    每次二分答案,把枚举两个点距离(>mid)的就连边限制
    傻逼到没输出-1,WA无数遍

    # include <iostream>
    # include <stdio.h>
    # include <stdlib.h>
    # include <string.h>
    # include <math.h>
    # include <algorithm>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(1005);
    const int __(4e6 + 5);
    
    IL int Input(){
        RG int x = 0, z = 1; RG char c = getchar();
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    int n, A, B, tmp, first[_], cnt, num, mx, mn = 2e9;
    int x[_], y[_], X1, Y1, X2, Y2, d1[_], d2[_], dis;
    int S[_], vis[_], dfn[_], low[_], Index, col[_];
    struct Link{
    	int u, v;
    } hate[_], like[_];
    struct Edge{
    	int to, next;
    } edge[__];
    
    IL void Add(RG int u, RG int v){
    	edge[cnt] = (Edge){v, first[u]}; first[u] = cnt++;
    }
    
    IL void Tarjan(RG int u){
    	vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;
    	for(RG int e = first[u]; e != -1; e = edge[e].next){
    		RG int v = edge[e].to;
    		if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);
    		else if(vis[v]) low[u] = min(low[u], dfn[v]);
    	}
    	if(dfn[u] != low[u]) return;
    	RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;
    	while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;
    }
    
    IL int Calc(RG int a, RG int b, RG int c, RG int d){
    	return abs(a - c) + abs(b - d);
    }
    
    IL int Check(RG int mid){
    	Fill(first, -1), Fill(dfn, 0), Fill(col, 0), cnt = Index = num = 0;
    	for(RG int i = 1; i <= A; ++i){
    		Add(hate[i].u, hate[i].v + n), Add(hate[i].u + n, hate[i].v);
    		Add(hate[i].v, hate[i].u + n), Add(hate[i].v + n, hate[i].u);
    	}
    	for(RG int i = 1; i <= B; ++i){
    		Add(like[i].u, like[i].v), Add(like[i].u + n, like[i].v + n);
    		Add(like[i].v, like[i].u), Add(like[i].v + n, like[i].u + n);
    	}
    	for(RG int i = 1; i < n; ++i)
    		for(RG int j = i + 1; j <= n; ++j){
    			if(d1[i] + d1[j] > mid) Add(i, j + n), Add(j, i + n);
    			if(d2[i] + d2[j] > mid) Add(i + n, j), Add(j + n, i);
    			if(d1[i] + dis + d2[j] > mid) Add(i, j), Add(j + n, i + n);
    			if(d2[i] + dis + d1[j] > mid) Add(i + n, j + n), Add(j, i);
    		}
    	for(RG int tmp = n << 1, i = 1; i <= tmp; ++i) if(!dfn[i]) Tarjan(i);
    	for(RG int i = 1; i <= n; ++i) if(col[i] == col[i + n]) return 0;
    	return 1;
    }
    
    int main(RG int argc, RG char* argv[]){
    	n = Input(), A = Input(), B = Input();
    	X1 = Input(), Y1 = Input(), X2 = Input(), Y2 = Input();
    	dis = Calc(X1, Y1, X2, Y2);
    	for(RG int i = 1; i <= n; ++i){
    		x[i] = Input(), y[i] = Input();
    		d1[i] = Calc(x[i], y[i], X1, Y1);
    		d2[i] = Calc(x[i], y[i], X2, Y2);
    		mx = max(mx, max(d1[i], d2[i]));
    		mn = min(mn, min(d1[i], d2[i]));
    	}
    	for(RG int i = 1; i <= A; ++i) hate[i] = (Link){Input(), Input()};
    	for(RG int i = 1; i <= B; ++i) like[i] = (Link){Input(), Input()};
    	mx = mx * 2 + dis, mn *= 2;
    	RG int l = mn, r = mx, ans = -1;
    	while(l <= r){
    		RG int mid = (l + r) >> 1;
    		if(Check(mid)) ans = mid, r = mid - 1;
    		else l = mid + 1;
    	}
    	printf("%d
    ", ans);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8463201.html
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