BZOJ4326: NOIP2015 运输计划

从题解里看到了二分才想到二分

题意为：给出一棵树和一些路线，要求你选树上一条边边权变为0后，路线中最大值最小

显然答案有单调性，二分答案

将所有长度超过 mid 的路径在树上标记出来，找出一条被所有的标记路径覆盖的边，当前答案合法条件为：max_path - maxlen <= mid

显然上边的操作可以树上差分，预处理每条路径端点的 lca 即可做到 O(n) 判断

这个 O(n) 大概是 2n ，卡卡常就好了

---

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cstdio>
using namespace std;

const int MAXN = 300005;

struct EDGE{
  int nxt, to, val;
  EDGE(int NXT = 0, int TO = 0, int VAL = 0) {nxt = NXT; to = TO; val = VAL;}
}edge[MAXN << 1];
struct QUE{
  int x, y, lca, dst;
}que[MAXN];
int n, m, totedge, mxl, mnl = 0x7fffffff;
int head[MAXN], dst[MAXN], tag[MAXN], top[MAXN];
int dep[MAXN], siz[MAXN], son[MAXN], fa[MAXN];

inline int rd() {
  register int x = 0;
  register char c = getchar();
  while(!isdigit(c)) c = getchar();
  while(isdigit(c)) {
    x = x * 10 + (c ^ 48);
    c = getchar();
  }
  return x;
}
inline int mx(int x, int y) {
  return (x > y ? x : y);
}
inline int mn(int x, int y) {
  return (x > y ? y : x);
}
void gfs(int x) {
  siz[x] = 1;
  for(int i = head[x]; i; i = edge[i].nxt) if(edge[i].to != fa[x]) {
    int y = edge[i].to;
    dep[y] = dep[x] + 1;
    dst[y] = dst[x] + edge[i].val;
    fa[y] = x;
    gfs(y);
    siz[x] += siz[y];
    if(siz[y] > siz[son[x]]) son[x] = y;
  }
  return;
}
void ifs(int x) {
  if(!top[x]) top[x] = x;
  if(son[x]) {
    top[son[x]] = top[x];
    ifs(son[x]);
  }
  for(int i = head[x]; i; i = edge[i].nxt) if(edge[i].to != son[x] && edge[i].to != fa[x]) ifs(edge[i].to);
  return;
}
inline int lca(int x, int y) {
  while(top[x] != top[y]) {
    if(dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  return (dep[x] > dep[y] ? y : x);
}
int dfs(int x, int tot, int frm) {
  int maxlen = 0;
  for(int i = head[x]; i; i = edge[i].nxt) if(fa[x] != edge[i].to) {
      maxlen = max(maxlen, dfs(edge[i].to, tot, i));
      tag[x] += tag[edge[i].to];
  }
  return (tag[x] == tot) ? (mx(maxlen, edge[frm].val)) : maxlen;
}
inline bool chk(int mid) {
  register int tot = 0, maxlen = 0;
  register bool extra = false;
  for(int i = 1; i <= n; ++i) tag[i] = 0;
  for(int i = 1; i <= m; ++i) {
    if(que[i].dst > mid) {
      maxlen = mx(maxlen,que[i].dst);
      ++tot;
      ++tag[que[i].x]; ++tag[que[i].y];
      tag[que[i].lca] -= 2;
      extra = true;
    }
  }
  if(!extra) return true;
  return (maxlen - dfs(1, tot, 0) <= mid);
}
inline void hfs(int l, int r) {
  int mid = 0, ans = 0;
  while(l < r) {
    mid = ((l + r) >> 1);
    if(chk(mid)) {
      ans = mid;
      r = mid;
    }
    else l = mid + 1;
  }
  printf("%d\n", ans);
  return;
}

int main() {
  //  freopen("testdata1.in", "r", stdin);
  n = rd(); m = rd();
  register int xx, yy, vv, tmp, N;
  tmp = (n - 1) & 1; N = n - 1 - tmp;
 //printf("tmp = %d, N = %d\n", tmp, N);
  for(int i = 1; i <= N; i += 2) {
  //	printf("i = %d\n", i);
  //	printf("totedge = %d\n", totedge);
    xx = rd(); yy = rd(); vv = rd();
    mnl = mn(mnl,vv);
	edge[++totedge] = EDGE(head[xx], yy, vv);
	head[xx] = totedge;
	edge[++totedge] = EDGE(head[yy], xx, vv);
	head[yy] = totedge;
    xx = rd(); yy = rd(); vv = rd();
    mnl = mn(mnl,vv);
	edge[++totedge] = EDGE(head[xx], yy, vv);
	head[xx] = totedge;
	edge[++totedge] = EDGE(head[yy], xx, vv);
	head[yy] = totedge;
//	printf("totedge = %d\n", totedge);
  }
  for(int i = 1; i <= tmp; ++i) {
    xx = rd(); yy = rd(); vv = rd();
    mnl = mn(mnl,vv);
	edge[++totedge] = EDGE(head[xx], yy, vv);
	head[xx] = totedge;
	edge[++totedge] = EDGE(head[yy], xx, vv);
	head[yy] = totedge;
  }
  dst[1] = 0; dep[1] = 1;
  gfs(1);
  ifs(1);
  for(int i = 1; i <= m; ++i) {
    que[i].x = rd(); que[i].y = rd();
    que[i].lca = lca(que[i].x, que[i].y);
    que[i].dst = dst[que[i].x] + dst[que[i].y] - (dst[que[i].lca] << 1);
    mxl = mx(mxl,que[i].dst);
  }
  hfs(mnl, mxl);
  return 0;
}