【BZOJ4326】NOIP2015 运输计划【LCA】【路径求并】

来源：互联网发布：淘宝买家提取器编辑：程序博客网时间：2024/05/18 00:01

【题目链接】

大概并没有用正解，卡常数A了。

先二分答案mid，那么边权变为0的边一定在所有长度大于mid的路径的交上，且这条边的边权至少为路径长度减去mid（否则不可能长度不可能降低到mid之下）。

每次check，把长度大于mid的路径加进去，求一次交（一次dfs就求出来了），遍历每条边，看是否有边满足上述条件。

/* Pigonometry */#include <cstdio>#include <algorithm>using namespace std;const int maxn = 300005, maxm = 300005, maxk = 20;int n, m, head[maxn], cnt;struct _edge {int v, w, next;} g[maxn << 1];struct _line {int u, v, lca, dis;} line[maxm];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void add(int u, int v, int w) {g[cnt] = (_edge){v, w, head[u]};head[u] = cnt++;}/* doubling lca */int pre[maxn][maxk], val[maxn], depth[maxn], dis[maxn];inline void dfs(int x) {for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ pre[x][0]) {pre[g[i].v][0] = x;val[g[i].v] = g[i].w;depth[g[i].v] = depth[x] + 1;dis[g[i].v] = dis[x] + g[i].w;dfs(g[i].v);}}inline int getlca(int u, int v) {if(depth[u] < depth[v]) swap(u, v);for(int i = maxk - 1; i >= 0; i--) if(depth[pre[u][i]] >= depth[v]) u = pre[u][i];for(int i = maxk - 1; i >= 0; i--) if(pre[u][i] != pre[v][i]) u = pre[u][i], v = pre[v][i];return u == v ? u : pre[u][0];}int sum[maxn];inline void work(int x) {for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ pre[x][0]) {work(g[i].v);sum[x] += sum[g[i].v];}}inline bool check(int x) {int tot = 0, w = 0;for(int i = 1; i <= n; i++) sum[i] = 0;for(int i = 1; i <= m; i++) if(line[i].dis > x) {tot++;sum[line[i].u]++;sum[line[i].v]++;sum[line[i].lca] -= 2;w = max(w, line[i].dis - x);}work(1);for(int i = 1; i <= n; i++)if(sum[i] == tot && val[i] >= w) return 1;return 0;}int main() {n = iread(); m = iread();for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0;for(int i = 1; i < n; i++) {int u = iread(), v = iread(), w = iread();add(u, v, w); add(v, u, w);}dfs(1);for(int j = 1; j < maxk; j++) for(int i = 1; i <= n; i++)pre[i][j] = pre[pre[i][j - 1]][j - 1];int l = 0, r = 0;for(int i = 1; i <= m; i++) {int u = iread(), v = iread(), lca = getlca(u, v), d = dis[u] + dis[v] - 2 * dis[lca];line[i] = (_line){u, v, lca, d};r = max(r, d);}while(l <= r) {int mid = l + r >> 1;if(check(mid)) r = mid - 1;else l = mid + 1;}printf("%d\n", l);return 0;}

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