标签：4条 ast string cto line algorithm last ble long
嘟嘟嘟
最近把21天漏的给不上。
今天重温了一下2-SAT，感觉很简单。就是把所有条件都转化成如果……必然能导出……。然后就这样连边建图，这样一个强连通分量中的所有点必然都是真或者假。从而根据这个点拆点后的两个点是否在一个强连通分量里判断是否有解。
这题人很容易想到拆点：\(i\)表示\(i\)连向\(s_1\)，\(i + n\)表示\(i\)连向\(s_2\)。
这道题关键在于距离这个限制。我们肯定能想到二分，但是接下来我就没想出来，因为2-SAT的图的边权是没有意义的。但实际上这个也是可以用2-SAT的思路解决的：比如\(dis(i, s_1) + dis(j, s_1) > mid\)，那么就说明，如果\(i\)连了\(s_1\)，\(j\)只能连\(s_2\)，因此我们连边\((i, j + n)\)。按照这种思路把四种情况都写一遍就行了。
这题和板子还有一个区别就是，板子给的条件形如"如果\(i\)为真，则\(j\)为假"，也就是说，每一点的真假已知。但这到题真假不定，所以都得讨论。也就是说，对于每一个讨厌和喜欢关系，都得连4条边。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e2 + 5;
const int maxe = 5e6 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, A, B;
struct Node
{
int x, y;
friend In int dis(Node& A, Node& B)
{
return abs(A.x - B.x) + abs(A.y - B.y);
}
}s[2], p[maxn], a[maxn << 1], b[maxn << 1];
int d[maxn << 1], DIS;
struct Edge
{
int nxt, to;
}e[maxe];
int head[maxn << 1], ecnt = -1;
In void addEdge(int x, int y)
{
e[++ecnt] = (Edge){head[x], y};
head[x] = ecnt;
}
In void build(int x)
{
for(int i = 1; i <= A; ++i)
{
int u = a[i].x, v = a[i].y;
addEdge(u, v + n), addEdge(v, u + n);
addEdge(u + n, v), addEdge(v + n, u);
}
for(int i = 1; i <= B; ++i)
{
int u = b[i].x, v = b[i].y;
addEdge(u, v), addEdge(u + n, v + n);
addEdge(v, u), addEdge(v + n, u + n);
}
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
if(i ^ j)
{
if(d[i] + d[j] > x)
addEdge(i, j + n), addEdge(j, i + n);
if(d[i + n] + d[j + n] > x)
addEdge(i + n, j), addEdge(j + n, i);
if(d[i] + d[j + n] + DIS > x)
addEdge(i, j), addEdge(j + n, i + n);
if(d[i + n] + d[j] + DIS > x)
addEdge(i + n, j + n), addEdge(j, i);
}
}
bool in[maxn << 1];
int dfn[maxn << 1], low[maxn << 1], cnt = 0;
int st[maxn << 1], top = 0, col[maxn << 1], ccol = 0;
In void dfs(int now)
{
dfn[now] = low[now] = ++cnt;
st[++top] = now; in[now] = 1;
for(int i = head[now], v; ~i; i = e[i].nxt)
{
if(!dfn[v = e[i].to])
{
dfs(v);
low[now] = min(low[now], low[v]);
}
else if(in[v]) low[now] = min(low[now], dfn[v]);
}
if(low[now] == dfn[now])
{
int x; ++ccol;
do
{
x = st[top--]; in[x] = 0;
col[x] = ccol;
}while(x ^ now);
}
}
In void init()
{
ecnt = -1;
for(int i = 1; i <= (n << 1); ++i)
dfn[i] = low[i] = col[i] = in[i] = 0, head[i] = -1;
cnt = top = ccol = 0;
}
In bool sat(int x)
{
init();
build(x);
for(int i = 1; i <= (n << 1); ++i) if(!dfn[i]) dfs(i);
for(int i = 1; i <= n; ++i)
if(col[i] == col[i + n]) return 0;
return 1;
}
int main()
{
n = read(), A = read(), B = read();
s[0].x = read(), s[0].y = read(), s[1].x = read(), s[1].y = read();
for(int i = 1; i <= n; ++i) p[i].x = read(), p[i].y = read();
for(int i = 1; i <= A; ++i) a[i].x = read(), a[i].y = read();
for(int i = 1; i <= B; ++i) b[i].x = read(), b[i].y = read();
DIS = dis(s[0], s[1]);
for(int i = 1; i <= n; ++i) d[i] = dis(p[i], s[0]), d[i + n] = dis(p[i], s[1]);
int L = 0, R = 6e6;
while(L < R)
{
int mid = (L + R) >> 1;
if(sat(mid)) R = mid;
else L = mid + 1;
}
write(L == 6e6 ? -1 : L), enter;
return 0;
}标签：4条 ast string cto line algorithm last ble long
原文地址：https://www.cnblogs.com/mrclr/p/10741117.html
