【BZOJ4078】[Wf2014]Metal Processing Plant【2-SAT】【二分】【二分圖】【並查集】

【題目鏈接】

考慮比較暴力的方法，我們枚舉兩個集合的最大值S1, S2，那麼我們可以用2-SAT來判斷合法不合法（如果i, j之間的值大於S1，那麼如果i在第一個集合，j只能在第二個集合，其他類似）。

我們將邊權從大到小排序，依次枚舉S1，發現S2是單調的（S2越大，越可能合法），於是可以二分S2了。

另外還有個優化，把枚舉S1的過程看成加邊的過程，我們發現當這個圖不是二分圖的時候就可以停止枚舉了。

令m = n * n，C爲最大邊權，複雜度爲O(mlogm + 玄學*(m + n)logC)

/* Think Thank Thunk */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 205, maxm = maxn << 1, inf = 0x3f3f3f3f;

int n, m, A[maxn][maxn], fa[maxn], d[maxn];
int head[maxm], cnt, dfn[maxm], low[maxm], belong[maxm], tot, clo, sta[maxm], top;
bool ins[maxm];

struct edge {
	int u, v, w;

	bool operator < (const edge &x) const {
		return w != x.w ? w > x.w : u != x.u ? u > x.u : v > x.v;
	}
} e[maxn * maxn];

struct _edge {
	int v, next;
} g[maxm * maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v) {
	g[cnt] = (_edge){v, head[u]};
	head[u] = cnt++;
}

inline int find(int x) {
	if(fa[x] == x) return x;
	int res = find(fa[x]);
	d[x] ^= d[fa[x]];
	return fa[x] = res;
}

inline void tarjan(int x) {
	dfn[x] = low[x] = ++clo;
	ins[sta[++top] = x] = 1;
	for(int i = head[x]; ~i; i = g[i].next) {
		int v = g[i].v;
		if(!dfn[v]) tarjan(v), low[x] = min(low[x], low[v]);
		else if(ins[v]) low[x] = min(low[x], dfn[v]);
	}
	if(dfn[x] == low[x]) {
		tot++;
		while(1) {
			int u = sta[top--];
			belong[u] = tot;
			ins[u] = 0;
			if(u == x) break;
		}
	}
}

inline bool check(int S1, int S2) {
	for(int i = n << 1; i; i--) head[i] = -1, dfn[i] = low[i] = belong[i] = ins[i] = 0; cnt = clo = tot = 0;
	for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) {
		if(A[i][j] > S1) {
			add((i << 1) - 1, j << 1);
			add((j << 1) - 1, i << 1);
		}
		if(A[i][j] > S2) {
			add(i << 1, (j << 1) - 1);
			add(j << 1, (i << 1) - 1);
		}
	}
	for(int i = n << 1; i; i--) if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= n; i++) if(belong[(i << 1) - 1] == belong[i << 1]) return 0;
	return 1;
}

inline int calc(int c) {
	int l = 0, r = c;
	while(l <= r) {
		int mid = l + r >> 1;
		if(check(c, mid)) r = mid - 1;
		else l = mid + 1;
	}
	return l;
}

int main() {
	n = iread();
	if(n <= 2) {
		printf("0\n");
		return 0;
	}
	for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) {
		A[i][j] = A[j][i] = iread();
		e[++m] = (edge){i, j, A[i][j]};
	}
	for(int i = 1; i <= n; i++) fa[i] = i, d[i] = 0;
	sort(e + 1, e + m + 1);

	int ans = inf;
	for(int i = 1; i <= m; i++) {
		int u = e[i].u, v = e[i].v, w = e[i].w;
		if(find(u) != find(v)) {
			int x = find(u), y = find(v);
			ans = min(ans, w + calc(w));
			d[x] = d[u] ^ d[v] ^ 1;
			fa[x] = y;
		} else if(d[u] == d[v]) {
			ans = min(ans, w + calc(w));
			break;
		}
	}
	printf("%d\n", ans);
	return 0;
}