Educational Codeforces Round 54 (Rated for Div.2)

D. Edge Deletion

题意：一张n个点的无向图，保留其中k条边，使得有尽可能多的点与1的最短路长度不变。

做法：求出最短路树，然后自底向上删边即可。

#include <bits/stdc++.h>
#define pb push_back
#define P pair<ll,int>
typedef long long ll;
const ll inf = 1e18;
const int N = 3e5 + 7;
using namespace std;
int n, m , k;
struct edge{int e,nxt,id; ll w;
}E[N<<1],E2[N<<1];
int h[N], cc, h2[N],cc1;
void add(int u,int v,ll w,int d) {E[cc].e = v; E[cc].w = w; E[cc].id = d;E[cc].nxt = h[u]; h[u] = cc; ++cc;
}
void add2(int u,int v,ll w,int d) {E2[cc1].e = v; E2[cc1].w = w; E2[cc1].id = d;E2[cc1].nxt = h2[u]; h2[u] = cc1; ++cc1;
}
struct node{int x;ll d;node(){}node(int a,ll b){x=a;d=b;}bool operator < (const node a)const {return a.d < d;}
};
ll dis[N];
int vis[N], fa[N], fr[N];
void dij() {for(int i=1;i<=n;++i)dis[i]=inf;priority_queue<node> q;q.push(node(1,0));dis[1]=0; fa[1] = 0; fr[1] = -1;while(!q.empty()) {node tmp = q.top(); q.pop();int u=tmp.x;if(vis[u])continue;vis[u]=1;for(int i=h[u];~i;i=E[i].nxt) {int v=E[i].e;if(dis[v]>dis[u]+E[i].w) {dis[v]=dis[u]+E[i].w;fa[v] = u;fr[v] = E[i].id;q.push(node(v,dis[v]));}}}return;
}
struct node2{int u,v,id; ll w;node2(){}node2(int a,int b,ll c, int d) {u=a; v = b; w = c; id = d;}
};
node2 A[N];int dep[N];
void bfs() {queue<int> q;memset(dep,-1,sizeof(dep));q.push(1); dep[1] = 0;while(!q.empty()) {int u = q.front(); q.pop();for(int i = h2[u]; ~i ; i = E2[i].nxt) {int v = E2[i].e;if(dep[v] == -1) {dep[v] = dep[u] + 1;q.push(v);}}}
}vector< P > B;
int vis2[N];
int main() {scanf("%d%d%d",&n,&m,&k);memset(h,-1,sizeof(h));memset(h2,-1,sizeof(h2));for(int i = 1; i <= m; ++i) { int u,v; ll w;scanf("%d%d%lld",&u,&v,&w);A[i] = node2(u,v,w,i);add(u,v,w,i); add(v,u,w,i);}dij();for(int i = 2; i <= n; ++i) {int p = fr[i];vis2[p] = 1;add2(A[p].u,A[p].v,A[p].w,A[p].id);add2(A[p].v,A[p].u,A[p].w,A[p].id);}int e = n-1;bfs();for(int i = 2; i <= n; ++i) B.pb(P(dep[i],i));sort(B.begin(),B.end());for(int i = (int)B.size()-1; i >= 0; --i) {if(e > k) {vis2[fr[B[i].second]] = 0;--e;}}printf("%d\n",e);for(int i = 1; i <= m; ++i) if(vis2[i]) printf("%d ",i); puts("");
}

E. Vasya and a Tree

题意：给定一颗树，进行m个操作，每次将节点v子树中向下d+1层，的点全部加x，操作完成后询问每个点的值。

做法：dfs这棵树的同时，树状数组维护对应深度的影响，退出递归时，还原现场即可，类似于树上逆序对，因为操作的总和为m所以复杂度有保证。kd-tree和二维树状数组，都没卡过去。。。

#include <bits/stdc++.h>
#define pb push_back
#define fr first
#define sc second
#define P pair<int,ll>
typedef long long ll;
const int N = 300100;
using namespace std;
int n, m;
int dep[N],MX;
vector<int> G[N];
vector< P > A[N];
ll B[N], ans[N<<1];
void add(int x,ll v) {x += 10;for(int i=x;i;i-=(i&-i)) B[i] += v;
}
ll ask(int x) {ll ans = 0;x += 10;for(int i = x; i <= MX+20; i+=(i&(-i))) ans += B[i];return ans;
}
void dfs(int u,int fa) {dep[u] = dep[fa] + 1;MX = max(dep[u],MX);for(int i = 0; i < G[u].size(); ++i) {int v = G[u][i];if(v != fa) dfs(v,u);}
}
void dfs2(int u,int fa) {for(int i = 0; i < A[u].size(); ++i) add(A[u][i].fr,A[u][i].sc);ans[u] = ask(dep[u]);for(int i = 0; i < G[u].size(); ++i) {int v = G[u][i];if(v != fa) {dfs2(v,u);}}for(int i = 0; i < A[u].size(); ++i) add(A[u][i].fr,-A[u][i].sc);
}
int main() {scanf("%d",&n);for(int i = 1; i <= n-1; ++i) { int u,v;scanf("%d%d",&u,&v);G[u].pb(v); G[v].pb(u);}dep[0] = -1;dfs(1,0);scanf("%d",&m);for(int i = 1; i <= m; ++i) { int v,d; ll x;scanf("%d%d%lld",&v,&d,&x);A[v].pb(P(min(dep[v]+d,MX),x));}dfs2(1,0);for(int i = 1; i <= n; ++i)printf("%lld ",ans[i]);puts("");return 0;
}

F. Summer Practice Report

题意：有\(n\)页纸，第\(i\)页包含\(a[i]\)个\(T\), \(b[i]\)个\(F\)，要求将所有的\(n\)页纸并起来后，不能有连续的\(k\)个\(T\)或\(F\)，问是否有解。

做法：贪心构造dp。\(dp[i][0/1]\) 表示前\(i\)页纸放完，最后几个字符是\(T\)或\(F\)时，\(T\)或\(F\)最小的数目。如果\(min(dp[n][0], dp[n][1]) <= k\) 则满足条件。考虑如何\(dp\)，设上一张末尾的\(T\)有\(pa\)张或\(F\)有\(pb\)张，当前这一张有\(a\)个\(T\)，\(b\)个\(F\)。
先确定\(dp[i][0]\)的转移，考虑放满\(T\)然后向其中插入\(F\)，用\(pa\)更新答案，那么如果\(pa<=k\)时，\(num\)即是需要插入的最少的\(F\)的个数，如果\(b == num\)， 那么用最后剩下的\(T\)更新答案，同时可以知道\(b\)的上界就是每个\(T\)之间都插入\(k\)个\(F\)，如果\(b>num\)且\(b <= k*a\)，就可以在最后一个\(T\)之前插入一个\(F\)，使得答案为\(1\)。用\(pb\)更新答案，思路类似，需要修改一下限制条件。\(dp[i][1]\) 也可以同样的转移。注意过程中会爆\(int\)

这道题在\(dp\)的同时贪心的转移，感觉思路十分清奇，看懂官方题解感觉自己dp烂的不要不要的。。。

#include <bits/stdc++.h>
typedef long long ll;
const int N = 300000+5;
const ll inf = 0x3f3f3f3f3f3f3f3f3f3f;
using namespace std;
int n, k, a[N], b[N];
int dp[N][2];
int cal(int pa, int pb, int a, int b) {ll ans = inf;if(pa <= k) {int num = (a + pa) / k + !!((a + pa) % k) - 1;if(b == num)ans = min(ans, pa + a - (ll)num*k);else if(b > num && (ll)b <= (ll)a*k)ans = min(ans, 1ll);}if(pb <= k) {int num = a / k + !!(a % k) - 1;if(b == num)ans = min(ans, a - (ll)num*k);else if(b > num && (ll)b <= (ll)(a-1)*k + (k-pb))ans = min(ans, 1ll);}return (int)ans;
}
int main() {scanf("%d %d", &n, &k);for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);for(int i = 1; i <= n; ++i) scanf("%d", &b[i]);for(int i = 1; i <= n; ++i) dp[i][0] = dp[i][1] = inf;for(int i = 1; i <= n; ++i) {dp[i][0] = cal(dp[i-1][0], dp[i-1][1], a[i], b[i]);dp[i][1] = cal(dp[i-1][1], dp[i-1][0], b[i], a[i]);}if(dp[n][0] <= k || dp[n][1] <= k) puts("YES");else puts("NO");return 0;
}