loj#2718. 「NOI2018」归程

题目链接

loj#2718. 「NOI2018」归程

题解

按照高度做克鲁斯卡尔重构树

那么对于询问倍增找到当前点能到达的高度最小可行点，该点的子树就是能到达的联通快，维护子树中到1节点的最短距离

spfa她死了...同步赛没写的说...

似乎前两题比去年简单些....连蒟蒻我都可做前两题的说

代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c <= '9' && c >= '0')x = x * 10 + c - '0',c = getchar();
	return x * f;
}
#define mp std::make_pair
#define pr std::pair<int,int>
int n,m;
const int maxn = 1000007;
struct node {
	int u,v,a,l,next;
	bool operator  < (const node & A)const {
		return a > A.a;
	}
} edge[maxn << 1],e[maxn]; 

int num = 0,head[maxn],dis[maxn << 1];
inline void add_edge(int u,int v,int w) { edge[++ num].v = v;edge[num].l = w;edge[num].next = head[u];head[u] = num;  }
std::priority_queue<pr> q;
void dijkstar() {
	static bool vis[maxn];
	memset(dis,0x3f,sizeof dis); memset(vis,0,sizeof vis);
	dis[1] = 0;q.push(pr(0,1));
	while(!q.empty()) {
		int u = q.top().second;q.pop();
		if(vis[u]) continue;
		vis[u] = 1;
		for(int i = head[u];i;i = edge[i].next) {
			int v = edge[i].v;
			if(dis[v] > dis[u] + edge[i].l) dis[v] = dis[u] + edge[i].l,q.push(mp(-dis[v],v));
		}
	}
}
int fa[maxn];
int find(int x) { if(fa[x] != x) fa[x] = find(fa[x]); return fa[x];  }
int tot = 0,dad[maxn << 1][20],mn[maxn << 1];
int lastans = 0;
int query(int x,int p) {
	for(int i = 18;i >= 0;-- i) {
		if(mn[dad[x][i]] > p) x = dad[x][i];
	}
	return dis[x];
}
void solve() {
	int Q = read(),k = read(),S = read(),v,p;
	if(k) while(Q --) {
		v = (read() + k * lastans - 1) % n + 1;
		p = (read() + k * lastans ) % (S + 1);
		printf("%d\n",lastans = query(v,p)); 

	} else while(Q --) {
		v = read(),p = read();
		printf("%d\n",query(v,p));
	}
}
void init() {
	n = read(),m = read();
	//int cnt = 0;
	for(int u,v,a,l,i = 1;i <= m;++ i) {
		u = read(),v = read(),l = read(),a = read();
		e[i].u = u,e[i].v = v,e[i].a = a;
		add_edge(u,v,l); add_edge(v,u,l);
	}
	dijkstar();
	for(int i = 1;i <= n;++ i) fa[i] = i;
	tot = n;
	std::sort(e + 1,e + m + 1);
	int k = 1;
	for(int fx,fy,i = 1;i <= m;++ i) {
		int x = e[i].u,y = e[i].v,h = e[i].a;
		if((fx = find(x)) == (fy = find(y))) continue;
		fa[fx] = fa[fy] = dad[fx][0] = dad[fy][0] = ++ tot; fa[tot] = dad[tot][0] = tot;
		mn[tot] = e[i].a; dis[tot] = std::min(dis[fx],dis[fy]);
		if(++ k == n)break;
	}
	/*for(int i = 1;i <= tot;++ i)
		for(int j = 0;dad[i][j];++ j)
			dad[i][j + 1] = dad[dad[i][j]][j],mn[i][j + 1] = std::min(mn[i][j],mn[dad[i][j]][j]);
	*/
	for(int i = 1;i <= 18;++ i)
		for(int x = 1;x <= tot;++ x)dad[x][i] =  dad[dad[x][i - 1]][i - 1];
	solve();
}
int main() {
	freopen("return.in","r",stdin); freopen("return.out","w",stdout);
	int t = read();
	while(t --) {
		memset(head,0,sizeof head),num = 0; lastans = 0;
		init();
	}
	return 0;
}