[CF949C]Data Center Maintenance

題目大意：$n$個點，每個點有一個值$w_i$。$m$個條件，每個條件給出$x,y$，要求$w_x\not =w_y$。選擇最少的點，使其值加$1$後，所有條件成立（資料保證有解）。

題解：對于每個條件，若$(w_x+1)\bmod h=w_y$，連上$x->y$；若$(w_y+1)\bmod h=w_x$，連上$y->x$。一條邊的含義是，若起點加一，終點也要加一。縮點，強連通分量内的點要一起加。發現答案就是找最小的沒有出邊的點

卡點：無

C++ Code：

#include <cstdio>
#define maxn 100010
#define maxm maxn
#define gethour(x) ((x + 1) % h)
int n, m, h;
int w[maxn];

int head[maxn << 1], cnt;
struct Edge {
	int from, to, nxt;
} e[maxm << 2];
inline void addE(int a, int b) {
	e[++cnt] = (Edge) {a, b, head[a]}; head[a] = cnt;
}

int DFN[maxn], low[maxn], idx, sz[maxn];
int S[maxn], top, res[maxn], CNT;
bool ins[maxn];
inline int min(int a, int b) {return a < b ? a : b;}
void tarjan(int u) {
	DFN[u] = low[u] = ++idx;
	ins[S[++top] = u] = true;
	int v;
	for (int i = head[u]; i; i = e[i].nxt) {
		v = e[i].to;
		if (!DFN[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if (ins[v]) low[u] = min(low[u], DFN[v]);
	}
	if (DFN[u] == low[u]) {
		CNT++;
		do {
			ins[v = S[top--]] = false;
			sz[res[v] = CNT]++;
			addE(CNT + n, v);
		} while (u != v);
	}
}

int oud[maxn];
int main() {
	scanf("%d%d%d", &n, &m, &h);
	for (int i = 1; i <= n; i++) scanf("%d", w + i);
	for (int i = 0, a, b; i < m; i++) {
		scanf("%d%d", &a, &b);
		if (w[a] == gethour(w[b])) addE(b, a);
		if (w[b] == gethour(w[a])) addE(a, b);
	}
	int cnt_now = cnt;
	for (int i = 1; i <= n; i++) if (!DFN[i]) tarjan(i);
	for (int i = 1; i <= cnt_now; i++) {
		int u = e[i].from, v = e[i].to;
		if (res[u] != res[v]) oud[res[u]]++;
	}
	int ans = 0x3f3f3f3f, mini = n + 1;
	for (int i = 1; i <= CNT; i++) if (!oud[i] && ans > sz[i]) {
		ans = sz[i];
		mini = i + n;
	}
	printf("%d\n", ans);
	for (int i = head[mini]; i; i = e[i].nxt) printf("%d ", e[i].to);
	puts("");
	return 0;
}
           

　　

轉載于:https://www.cnblogs.com/Memory-of-winter/p/9745756.html