NOI2019同步赛 day2

今天的题目有些难

T1

一看是非常的可做
32分直接$O(n^2)$连边跑$dijkstra$
再来20分，使用二分查找对应点，然后连边，跑$dijkstra$
再来20分，我就不会啦，应该是线段树优化建边，我没学过，所以我在线学习了一下，然后是疯狂调试，看起来是过了……

满分的应该是想不出来的

Code：

#include 
#define maxn 100010
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
struct POS{
	int x, y, id;
}pos[maxn];
struct node{
	int val, len;
	bool operator < (const node &x) const{ return x.len < len; }
};
priority_queue <node> q;
int num, head[maxn], n, m, W, H, cnt;
int dis[maxn << 4], vis[maxn << 4];
int pp1[maxn], pp2[maxn];
struct Seg{
	int l, r, num;
}seg[5][maxn << 2];
struct Vec{
	int v, w;
};
vector <Vec> G[maxn << 4];

inline int read(){
    int s = 0, w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
    for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
    return s * w;
}

void addedge(int x, int y, int z){ edge[++num] = (Edge){ y, head[x], z }; head[x] = num; }

void dijkstra(){
	dis[1] = 0;
	for (int i = 2; i <= n; ++i) dis[i] = 1e9;
	q.push((node) { 1, 0 });
	while (!q.empty()){
		node tmp = q.top(); q.pop();
		int u = tmp.val, len = tmp.len;
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = head[u]; i; i = edge[i].next){
			int v = edge[i].to;
			if (dis[v] > len + edge[i].len){
				dis[v] = len + edge[i].len;
				if (!vis[v]) q.push( (node) { v, dis[v] });
			}
		}
	}
	for (int i = 2; i <= n; ++i) printf("%d\n", dis[i]);
}

void do1(){
	for (int i = 1; i <= m; ++i){
		int p = read(), t = read(), l = read(), r = read(), d = read(), u = read();
		for (int j = 1; j <= n; ++j)
			if (p != j && pos[j].x >= l && pos[j].x <= r && pos[j].y >= d && pos[j].y <= u)
				addedge(p, j, t);
	}
	dijkstra();
}

#define ls rt << 1
#define rs rt << 1 | 1 
typedef pair<int, int> pii;

void build(int rt, int l, int r, int op){
	seg[op][rt].l = l, seg[op][rt].r = r, seg[op][rt].num = ++cnt;
	if (l == r){
		if (op == 1) pp1[l] = cnt; else{
			pp2[l] = cnt;
			G[pp2[l]].push_back( (Vec) { pp1[l], 0} );
		}
		return;
	}
	int mid = (l + r) >> 1;
	build(ls, l, mid, op); build(rs, mid + 1, r, op);
	if (op == 1){
		G[seg[op][ls].num].push_back( (Vec){seg[op][rt].num, 0} );
        G[seg[op][rs].num].push_back( (Vec){seg[op][rt].num, 0} );
	} else{
        G[seg[op][rt].num].push_back( (Vec){seg[op][ls].num, 0} );
        G[seg[op][rt].num].push_back( (Vec){seg[op][rs].num, 0} );
	}
}

void update(int rt, int l, int r, int u, int w, int op){
	if(seg[op][rt].l == l && seg[op][rt].r == r) {
        if (op == 1) G[u].push_back( (Vec){seg[2][rt].num, w} ); else
		G[seg[1][rt].num].push_back( (Vec){u, w} );
        return;
    }
    int mid = (seg[op][rt].l + seg[op][rt].r) >> 1;
    if (r <= mid) update(ls, l, r, u, w, op); else
	if (l > mid) update(rs, l, r, u, w, op); else {
        update(ls, l, mid, u, w, op);
        update(rs, mid + 1, r, u, w, op);
    }
}

void do2_dijkstra(int s) {
    for (int i = 1; i <= cnt; ++i) dis[i] = 1e9;
    dis[pp1[s]] = 0;
    priority_queue<pii, vector<pii>, greater<pii> > q;
    q.push( (pii){0, pp1[s]} );
    while (!q.empty()) {
        int u = q.top().second; q.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        int v;
        for (int i = 0; i < (int) G[u].size(); ++i) {
            v = G[u][i].v;
            if (dis[v] > dis[u] + G[u][i].w) {
                dis[v] = dis[u] + G[u][i].w;
                q.push( (pii){dis[v], v} );
            }
        }
    }
}

void do2(){
	build(1, 1, W, 1); build(1, 1, W, 2);
	for (int i = 1; i <= m; ++i){
		int p = read(), t = read(), l = read(), r = read(), d = read(), u = read();
		update(1, l, r, pp1[pos[p].x], t, 1);
	}
	do2_dijkstra(pos[1].x); 
	for (int i = 2; i <= n; ++i) printf("%d\n", min(dis[pp1[pos[i].x]], dis[pp2[pos[i].x]]));
}

bool do3_cmp(POS a, POS b){ return a.x == b.x ? a.y < b.y : a.x < b.x; }

void do3(){
	for (int i = 1; i <= n; ++i) pos[i].id = i;
	sort(pos + 1, pos + 1 + n, do3_cmp);
	for (int i = 1; i <= m; ++i){
		int p = read(), t = read(), l = read(), r = read(), d = read(), u = read();
		int ll = 1, rr = n, q = 0;
		while (ll <= rr){
			int mid = (ll + rr) >> 1;
			if (pos[mid].x < l || pos[mid].x == l && pos[mid].y < d) ll = mid + 1; else
			rr = mid - 1, q = mid;
		}
		addedge(p, pos[q].id, t);
	}
	dijkstra();
}

int main(){
	freopen("jump.in", "r", stdin);
	freopen("jump.out", "w", stdout);
	n = read(), m = read(), W = read(), H = read();
	for (int i = 1; i <= n; ++i) pos[i].x = read(), pos[i].y = read();
	if (n <= 100) do1(); else
	if (H == 1) do2(); else do3(); 
	return 0;
}

T2

T1写了3.5h，后面的题目没时间写了，T2一看，没什么思路，直接上dfs拿10分

Code：

#include 
#define maxn 110
#define qy 998244353
#define int long long
using namespace std;
int ans[maxn], lena, lenb, a[maxn], b[maxn], f[maxn], A[maxn], n, m, type;

inline int read(){
    int s = 0, w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
    for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
    return s * w;
}

int ksm(int n, int k){
	if (!k) return 1;
	int sum = ksm(n, k >> 1);
	sum = 1LL * sum * sum % qy;
	if (k & 1) sum = 1LL * sum * n % qy;
	return sum;
}

void dfs(int i, int j, int p){
	int inv = ksm(i + j, qy - 2);
	(ans[i + j] += a[i] * i % qy * inv % qy * p % qy) %= qy;
	(ans[i + j] += b[j] * j % qy * inv % qy * p % qy) %= qy;
	if (i) dfs(i - 1, j, p * i % qy * inv % qy);
	if (j) dfs(i, j - 1, p * j % qy * inv % qy);
}

signed main(){
	freopen("landlords.in", "r", stdin);
	freopen("landlords.out", "w", stdout);
	n = read(), m = read(), type = read();
	for (int i = 1; i <= n; ++i) f[i] = type == 1 ? i : i * i;
	for (int i = 1; i <= m; ++i) A[i] = read();
	for (int i = 1; i <= A[1]; ++i) a[i] = f[i];
	for (int i = A[1] + 1; i <= n; ++i) b[i - A[1]] = f[i];
	lena = A[1], lenb = n - lena;
	dfs(lena, lenb, 1);
	int M = read();
	while (M--){
		int x = read();
		printf("%lld\n", ans[x]);
	}
	return 0;
}

T3

交互题什么的我完全不会了。。。
我不知道他是怎么操作的
所以这道题我直接弃疗了

所以期望得分72+10+0=82吧
如果能达到，两天加起来刚好200.。。。
感觉不错