NOIP2015 运输计划

BZOJ

CodeVS

Uoj

题目大意：

给一个n个点的边带权树，给定m条链，你可以选择树中的任意一条边，将它置为0，使得最长的链长最短。

题目分析：

最小化最大值，二分。

二分最短长度mid，将图中链长大于mid的链提取出来，求他们的交路径，选择他们都经过最大的一条边，看是否满足要求。
这是基本思路，下面来想想如何求解：一共尝试了两种办法：

• 倍增lca / 树链剖分lca + 树上差分： 求lca部分略过（本题用链剖较快），对于一条u->v的路径，在u处+1，v处+1，lca处-2，从下往上求后缀和，即可得到i->fa[i]这条边被经过的次数，假设二分时选出了k条边，那么选择一条经过次数==k的边权最大的边，来判断是否满足。

• 树链剖分 + 线段树：将u->v上每个节点+1，选出所有的边后，在线段树上将标记下放到底层（节点），如果该节点i（代表i->fa[i]这条边）值为k，更新边权最大值。最后用这个最大值来判断是否满足。注意:pathModify时，最后一步modify时不能改lca（看代码吧，一言难尽，仔细脑补），因为lca代表的边不是这条路径上的。

大优化： 开一个记忆化数组memory，记录选出最大的k条边时的最大权值和，可以加速很多。

ps: 被uoj的extra test hack掉了。

code

树链lca + 树上差分：

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

namespace IO{
    inline ll read(){
        ll i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline void wr(ll x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wr(x / 10);
        putchar(x % 10 + '0');
    }
}using namespace IO;

typedef long long ll;
const int N = 3e5 + 5, M = 3e5 + 5, OO = 0x3f3f3f3f;
int n, m, ecnt, adj[N], nxt[N << 1], go[N << 1];
int dep[N], sum[N], fa[N], sze[N], idx[N], pos[N], tot, son[N], top[N];
ll ans, len[N << 1], l = OO, r, dis[N], val[N], memory[M];
struct node{
    int u, v;
    ll tt;
    inline bool operator < (const node &b) const{return tt < b.tt;}
}plan[M];

inline void addEdge(int u, int v, ll t){nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = t;}

inline int getLca(int u, int v){
    while(top[u] != top[v]){
        if(dep[top[u]] < dep[top[v]]) swap(u, v);
        u = fa[top[u]];
    }
    if(u == v) return u;
    return dep[u] < dep[v] ? u : v;
}

inline void dfs1(int u, int f){
    dep[u] = dep[f] + 1, fa[u] = f, sze[u] = 1;
    for(int e = adj[u], v; e; e = nxt[e]){
        if((v = go[e]) == f) continue;
        dis[v] = dis[u] + len[e], val[v] = len[e], dfs1(v, u), sze[u] += sze[v];
        if(sze[v] > sze[son[u]]) son[u] = v;
    }
}

inline void dfs2(int u, int f){
    if(son[u]){
        idx[pos[son[u]] = ++tot] = son[u];
        top[son[u]] = top[u];
        dfs2(son[u], u);
    }
    for(int v, e = adj[u]; e; e = nxt[e]){
        if((v = go[e]) == f || v == son[u]) continue;
        top[v] = v;
        idx[pos[v] = ++tot] = v;
        dfs2(v, u);
    }
}

inline void getSum(int u, int f){
    for(int v, e = adj[u]; e; e = nxt[e]){
        if((v = go[e]) == f) continue;
        getSum(v, u);
        sum[u] += sum[v];
    }
}

inline bool check(ll mid){
    memset(sum, 0, sizeof sum); int cnt = 0;
    for(int i = m; i >= 1; i--){
        if(plan[i].tt <= mid) break;
        cnt++, sum[plan[i].u]++, sum[plan[i].v]++, sum[getLca(plan[i].u, plan[i].v)] -= 2;
    }
    if(memory[cnt])
        return plan[m].tt - memory[cnt] <= mid;
    getSum(1, 0); 
    ll maxx = 0;
    for(int i = 1; i <= n; i++) if(sum[i] == cnt) maxx = max(maxx, val[i]);
    memory[cnt] = maxx;
    return plan[m].tt - maxx <= mid;
}

inline void solve(){
    l = 0, r = plan[m].tt;
    while(l <= r){
        ll mid = l + r >> 1;
        if(check(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
}

int main(){
    n = read(), m = read();
    for(int i = 1; i < n; i++){
        int x = read(), y = read(); ll t = 1ll*read(); 
        addEdge(x, y, t), addEdge(y, x, t);
    } 
    pos[1] = top[1] = idx[1] = tot = 1, dfs1(1, 0), dfs2(1, 0);
    for(int i = 1; i <= m; i++) plan[i].u = read(), plan[i].v = read(), plan[i].tt = dis[plan[i].u] + dis[plan[i].v] - 2 * dis[getLca(plan[i].u, plan[i].v)];
    sort(plan + 1, plan + m + 1);
    solve();
    wr(ans), putchar('
');
    return 0;
}

树链剖分 + 线段树：

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

namespace IO{
    inline ll read(){
        ll i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline void wr(ll x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wr(x / 10);
        putchar(x % 10 + '0');
    }
}using namespace IO;

typedef long long ll;
const int N = 3e5 + 5, M = 3e5 + 5, OO = 0x3f3f3f3f;
int n, m, ecnt, adj[N], nxt[N << 1], go[N << 1];
int dep[N], sum[N], fa[N], sze[N], idx[N], pos[N], tot, son[N], top[N];
ll ans, len[N << 1], l = OO, r, dis[N], val[N], memory[M];
struct node{
    int u, v;
    ll tt;
    inline bool operator < (const node &b) const{return tt < b.tt;}
}plan[M];

namespace SegTree{
    int tree[N << 2], tag[N << 2];
    inline void upt(int k){tree[k] = tree[k << 1] + tree[k << 1 | 1];}
    inline void add(int k, int l, int r, int v){tree[k] += (r - l + 1) * v, tag[k] += v;}
    inline void modify(int k, int l, int r, int x, int y){
        if(x <= l && r <= y){add(k, l, r, 1);return;}
        int mid = l + r >> 1, lc = k << 1, rc = k << 1 | 1;
        if(x <= mid) modify(lc, l, mid, x, y);
        if(y > mid) modify(rc, mid + 1, r, x, y);
        upt(k);
    }
    inline void pushDown(int k, int l, int r){if(tag[k]) add(k << 1, l, l + r >> 1, tag[k]), add(k << 1 | 1, (l + r >> 1) + 1, r, tag[k]), tag[k] = 0;}
    inline void pushToTheEnd(int k, int l, int r, ll v, ll &maxx){
        if(l == r){if(tree[k] == v) maxx = max(maxx, val[idx[l]]);return;}
        int mid = l + r >> 1, lc = k << 1, rc = k << 1 | 1;
        pushDown(k, l, r);
        pushToTheEnd(lc, l, mid, v, maxx);
        pushToTheEnd(rc, mid + 1, r, v, maxx);
    }
}using namespace SegTree;

inline void addEdge(int u, int v, ll t){nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = t;}

inline int getLca(int u, int v){
    while(top[u] != top[v]){
        if(dep[top[u]] < dep[top[v]]) swap(u, v);
        u = fa[top[u]];
    }
    if(u == v) return u;
    return dep[u] < dep[v] ? u : v;
}

inline void dfs1(int u, int f){
    dep[u] = dep[f] + 1, fa[u] = f, sze[u] = 1;
    for(int e = adj[u], v; e; e = nxt[e]){
        if((v = go[e]) == f) continue;
        dis[v] = dis[u] + len[e], val[v] = len[e], dfs1(v, u), sze[u] += sze[v];
        if(sze[v] > sze[son[u]]) son[u] = v;
    }
}

inline void dfs2(int u, int f){
    if(son[u]){
        idx[pos[son[u]] = ++tot] = son[u];
        top[son[u]] = top[u];
        dfs2(son[u], u);
    }
    for(int v, e = adj[u]; e; e = nxt[e]){
        if((v = go[e]) == f || v == son[u]) continue;
        top[v] = v;
        idx[pos[v] = ++tot] = v;
        dfs2(v, u);
    }
}

inline void pathModify(int u, int v){
    while(top[u] != top[v]){
        if(dep[top[u]] < dep[top[v]]) swap(u, v);
        modify(1, 1, n, pos[top[u]], pos[u]);
        u = fa[top[u]];
    }
    if(dep[u] > dep[v]) swap(u, v);
    modify(1, 1, n, pos[son[u]], pos[v]);
}

inline bool check(ll mid){
    int cnt = 0;
    memset(tree, 0, sizeof tree);
    memset(tag, 0, sizeof tag);
    for(int i = m; i >= 1; i--){
        if(plan[i].tt <= mid) break;
        cnt++;
    }
    if(memory[cnt])
        return plan[m].tt - memory[cnt] <= mid; 
    for(int i = m; i >= m - cnt + 1; i--) pathModify(plan[i].u, plan[i].v);
    ll maxx = 0;
    pushToTheEnd(1, 1, n, cnt, maxx);
    memory[cnt] = maxx;
    return plan[m].tt - maxx <= mid;
}

inline void solve(){
    l = 0, r = plan[m].tt;
    while(l <= r){
        ll mid = l + r >> 1;
        if(check(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
}

int main(){
    n = read(), m = read();
    for(int i = 1; i < n; i++){
        int x = read(), y = read(); ll t = 1ll*read(); 
        addEdge(x, y, t), addEdge(y, x, t);
    } 
    pos[1] = top[1] = idx[1] = tot = 1, dfs1(1, 0), dfs2(1, 0);
    for(int i = 1; i <= m; i++) plan[i].u = read(), plan[i].v = read(), plan[i].tt = dis[plan[i].u] + dis[plan[i].v] - 2 * dis[getLca(plan[i].u, plan[i].v)];
    sort(plan + 1, plan + m + 1); 
    solve();
    wr(ans), putchar('
');
    return 0;
}