题目大意：$n$个点，每个点有一个值$w_i$。$m$个条件，每个条件给出$x,y$，要求$w_x\not =w_y$。选择最少的点，使其值加$1$后，所有条件成立（数据保证有解）。

题解：对于每个条件，若$(w_x+1)\bmod h=w_y$，连上$x->y$；若$(w_y+1)\bmod h=w_x$，连上$y->x$。一条边的含义是，若起点加一，终点也要加一。缩点，强连通分量内的点要一起加。发现答案就是找最小的没有出边的点

卡点：无

 

C++ Code：

#include 
     
      #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;}