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;

}

巴特西

loj#2718. 「NOI2018」归程

题目链接

题解

代码

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