Solution -「UR #2」「UOJ #32」跳蚤公路

\(\mathcal{Description}\)

  Link.

  给定一个 \(n\) 个点 \(m\) 条边的带权有向图，每条边还有属性 \(s\in\{-1,0,1\}\)。对于每个 \(u\in[1,n]\)，求有多少个 \(x\in\mathbb Z\)，使得图上所有属性为 \(-1\) 的边权 \(-x\)，为 \(0\) 的不变，为 \(1\) 的 \(+x\) 后，从 \(1\) 走到 \(u\) 的任意路径不经过负环。若存在无穷个 \(x\)，输出 \(-1\)。

  \(n\le100\)，\(m\le10^4\)。

\(\mathcal{Solution}\)

  可以发现，对于任意初始权和为 \(b\) 的简单环，设其上 \(\sum_us_u=k\)，则简单环的最终权值为 \(kx+b\)，是一个一次函数，我们需要保证它非负。

  难免涉及到找负环，首先排除 SPFA，我们可以利用 Bellman-Ford 找负环：令 \(f(i,u)\) 表示从起点走 \(i\) 步到达 \(u\) 的最短路，若 \(f(n-1,u)\le f(n,u)\)，则 \(u\) 和 \(u\) 可达的点都受负环影响。

  扩展到本题，我们需要记录 \(k\)，那么令 \(f(i,u,k)\) 表示从起点走 \(i\) 步到达 \(u\)，经过边的属性和为 \(k\) 的最短路，这个表可以暴力 \(\mathcal O(n^2m)\) 刷出来。对于一个确定的 \(x\)，若 \(u\) 被负环影响，联系上面二维情况下的式子，就有：

\[\min_j\{f(n-1,u,j)+jx\}\le\min_k\{f(n,u,k)+kx\} \]

  枚举 \(j,k\)，若 \(j<k\)，可以得到一个 \(x\) 的限制：

\[x\ge\frac{f(n-1,u,j)-f(n,u,k)}{k-j} \]

  其余情况类似。这样对于每个结点，我们可以求出关于其的若干限制区间 \((l,r)\)，表示合法的 \(x\not\in(l,r)\)。求答案时，枚举每个点 \(u\) 和其可达的点 \(v\)，求 \(v\) 上限制的交集即可。

\(\mathcal{Code}\)

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

typedef long long LL;
typedef std::pair<LL, LL> pll;

inline char fgc () {
	static char buf[1 << 17], *p = buf, *q = buf;
	return p == q && ( q = buf + fread ( p = buf, 1, 1 << 17, stdin ), p == q ) ? EOF : *p ++;
}

inline int rint () {
	int x = 0, f = 1; char s = fgc ();
	for ( ; s < '0' || '9' < s; s = fgc () ) f = s == '-' ? -f : f;
	for ( ; '0' <= s && s <= '9'; s = fgc () ) x = x * 10 + ( s ^ '0' );
	return x * f;
}

inline void wint ( LL x ) {
	if ( x < 0 ) putchar ( '-' ), x = -x;
	if ( 9 < x ) wint ( x / 10 );
	putchar ( x % 10 ^ '0' );
}

const int MAXN = 100, MAXM = 1e4;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, m;
LL pool[MAXN + 5][MAXN + 5][MAXN * 2 + 5];
bool rch[MAXN + 5][MAXN + 5];
std::vector<pll> restr[MAXN + 5], inter;

struct Edge {
	int u, v, w, s;
} eset[MAXM + 5];

inline void chkmin ( LL& a, const LL b ) { if ( b < a ) a = b; }
inline void chkmax ( LL& a, const LL b ) { if ( a < b ) a = b; }

inline LL& f ( const int i, const int j, const int k ) { return pool[i][j][k + n + 2]; }

int main () {
	// freopen ( "city.in", "r", stdin );
	// freopen ( "city.out", "w", stdout );
	n = rint (), m = rint ();
	for ( int i = 1; i <= n; ++ i ) rch[i][i] = true;
	for ( int i = 1, u, v, w, s; i <= m; ++ i ) {
		u = rint (), v = rint (), w = rint (), s = rint ();
		eset[i].u = u, eset[i].v = v, eset[i].w = w, eset[i].s = s;
		rch[u][v] = true;
	}
	for ( int k = 1; k <= n; ++ k ) {
		for ( int i = 1; i <= n; ++ i ) {
			for ( int j = 1; j <= n; ++ j ) {
				rch[i][j] |= rch[i][k] && rch[k][j];
			}
		}
	}
	for ( int j = 1; j <= n; ++ j ) {
		for ( int k = -n; k <= n; ++ k ) {
			f ( 0, j, k ) = INF;
		}
	}
	f ( 0, 1, 0 ) = 0;
	for ( int i = 1; i <= n; ++ i ) {
		for ( int j = 1; j <= n; ++ j ) {
			for ( int k = -n; k <= n; ++ k ) {
				f ( i, j, k ) = f ( i - 1, j, k );
			}
		}
		for ( int j = 1; j <= m; ++ j ) {
			int u = eset[j].u, v = eset[j].v, w = eset[j].w, s = eset[j].s;
			for ( int k = -n; k <= n; ++ k ) {
				if ( f ( i - 1, u, k ) == INF ) continue;
				chkmin ( f ( i, v, k + s ), f ( i - 1, u, k ) + w );
			}
		}
	}
	for ( int i = 1; i <= n; ++ i ) {
		for ( int k = -n; k <= n; ++ k ) {
			if ( f ( n, i, k ) >= f ( n - 1, i, k ) ) continue;
			LL l = -INF, r = INF; // x\in(-INF,l]+[r,INF).
			for ( int j = -n; j <= n; ++ j ) {
				if ( j == k || f ( n - 1, i, j ) == INF ) continue;
				if ( j < k ) {
					chkmin ( r, ceil ( 1.0 * ( f ( n - 1, i, j ) - f ( n, i, k ) ) / ( k - j ) ) );
				} else {
					chkmax ( l, floor ( 1.0 * ( f ( n - 1, i, j ) - f ( n, i, k ) ) / ( k - j ) ) );
				}
			}
			if ( l < r ) restr[i].push_back ( pll ( l, r ) );
		}
	}
	for ( int i = 1; i <= n; ++ i ) {
		inter.clear ();
		for ( int j = 1; j <= n; ++ j ) {
			if ( rch[1][j] && rch[j][i] ) {
				for ( int k = 0; k ^ restr[j].size (); ++ k ) {
					inter.push_back ( restr[j][k] );
				}
			}
		}
		if ( inter.empty () ) { puts ( "-1" ); continue; }
		std::sort ( inter.begin (), inter.end () );
		LL l = INF, r = -INF, las = -INF; bool found = false;
		for ( int j = 0; j ^ inter.size (); ++ j ) {
			if ( !j && inter[j].first > -INF ) {
				l = -INF, r = inter[j].first;
				found = true; break;
			}
			if ( las != -INF && las <= inter[j].first ) {
				l = las, r = inter[j].first;
				found = true; break;
			}
			chkmax ( las, inter[j].second );
		}
		if ( !found && las < INF ) l = las, r = INF;
		wint ( l == -INF || r == INF ? -1 : ( l <= r ? r - l + 1 : 0 ) );
		putchar ( '\n' );
	}
	return 0;
}

\(\mathcal{Details}\)

  关于负环的做得太少，拿着就不知所措 qwq，而且考场上前面浪费太多时间根本没来得及向这题……