[ 2020牛客NOIP赛前集训营-提高组（第六场） ] 补题
B.艰难睡眠
Problem
Solution
-
单调队列
-
段环成链思想
明显题目是要我们分别做 n 次单调队列，考虑如何写最方便。
思考后可以发现，题目中长度为 \(k\) 的睡眠段在移动，我们可以选择的数区间是一个长度为 \(m-k-b[i]+1\) 的区间。断环成链后，将跑一个长度为 \(m-k-b[i]+1\)，的区间的滑动窗口，若该区间的右端点为 \(r\)，则对应的睡眠段起点为 \(r+b[i]\)。
Code
Talk is cheap.Show me the code.
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
inline int read() {
int x = 0, f = 1; char ch = getchar();
while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
return x * f;
}
const int N = 5007, M = 2007;
int n,m,k;
int b[N],c[N][M],ans[M];
struct Node {
int val,t;
}q[M<<1];
int main()
{
//freopen("test.in","r",stdin);
n = read(), m = read(), k = read();
for(int i=1;i<=n;++i) {
int tmp = read(); b[i] = read();
}
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j) c[i][j] = read();
for(int i=1;i<=n;++i) {
if(b[i]+k > m) {
puts("-1"); return 0;
}
}
for(int t=1;t<=n;++t) {
int L = 1, R = 0;
int len = m - k - b[t] + 1;
for(int i=1;i<=2*m;++i) {
int now = c[t][(i-1) % m + 1];
while(L<=R && q[L].t<=i-len) ++L;
while(L<=R && q[R].val>=now) --R;
q[++R] = (Node)<%now, i%>;
if(i > m) ans[(i + b[t] - 1) % m + 1] += q[L].val;
}
}
int res = INF;
for(int i=1;i<=m;++i) res = min(res, ans[i]);
printf("%d\n",res);
return 0;
}
/*
3 10 6
3 1
2 1
2 3
4 4 8 5 1 2 2 2 6 4
9 3 7 5 7 1 5 2 2 8
8 2 7 7 4 9 2 3 4 3
5
*/
C.路径难题
Problem
Solution
对于每一路的公交车，建立一个虚拟节点，将该路公交车所有站点与虚拟节点相连，跑最短路。
用一个二元组 \((x,y)\) 表示路径长度，其中 \(y \in [0,r)\)。把边分为两类，一种乘公交车，一种不乘公交车。乘公交车要将 \(y\) 清空，如果 \(y>0\) 那么要将 \(x+1\) 再加上路径长度。
Code
Talk is cheap.Show me the code.
#include<bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
using namespace std;
inline int read() {
int x = 0, f = 1; char ch = getchar();
while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
return x * f;
}
typedef pair<int,int> PII;
const int N = 400007;
int n,m,cnt,K,r;
int head[N],c[N];
struct Edge {
int next,to,w,flag;
}edge[N<<1];
inline void add(int u,int v,int w,int flag) {
edge[++cnt] = (Edge)<%head[u],v,w,flag%>;
head[u] = cnt;
}
PII dist[N];
bool vis[N];
void Dijkstra(int s) {
memset(dist, 0x3f, sizeof(dist));
memset(vis, 0, sizeof(vis));
priority_queue<pair<PII,int> > q;
dist[s] = mp(0,0), vis[s] = 0; q.push(mp(dist[s], s));
while(!q.empty()) {
int u = q.top().se; vis[u] = 1; q.pop();
for(int i=head[u];i;i=edge[i].next) {
int v = edge[i].to, w = edge[i].w, flag = edge[i].flag;
PII now = dist[u];
if(flag == 0) {
now.se += w;
if(now.se >= r) {
now.fi += now.se/r, now.se %= r;
}
} else {
if(now.se > 0) now.fi++;
now.fi += w; now.se = 0;
}
//printf("%d->%d (%d,%d)->(%d,%d) %d %d\n",u,v,dist[u].fi,dist[u].se,now.fi,now.se,w,flag);
if(dist[v] > now) {
dist[v] = now;
now.fi = -now.fi; now.se = -now.se;
q.push(mp(now,v));
}
}
}
}
int main()
{
n = read(), m = read(), K = read(), r = read(); int q = read();
for(int i=1;i<=m;++i) {
int u = read(), v = read(), w = read();
add(u,v,w,0), add(v,u,w,0);
}
int tot = n;
for(int i=1;i<=K;++i) {
int t = read(); c[i] = read(); ++tot;
for(int j=1;j<=t;++j) {
int x = read(); add(x,tot,c[i],1), add(tot,x,0,0);
}
}
while(q--) {
int x = read(), y = read();
Dijkstra(x);
PII ans = dist[y];
if(ans.se > 0) ans.fi++;
printf("%d\n",ans.fi);
//PII now = dist[2];
//printf("%d %d\n",now.fi,now.se);
}
return 0;
}
/*
5 6 1 10 1
1 2 15
2 3 177
3 5 73
4 2 37
1 5 66
1 4 43
3 11 1 3 5
1 3
11
*/
