codeforces 337D D. Book of Evil (树形 dp)

来源：互联网发布：淘宝乔丹官方旗舰店编辑：程序博客网时间：2024/05/20 21:48

题解：我们发现可能藏着魔鬼的点就是那些，到所有被感染的点的最大距离小于等于d的点，剩下的我们的目的就是求出每个点到所有感染点的最大距离，这就是一个类似于求，树上的任意一个点，到所有点的最大距离，就是一个简单的树形dp;

#include <algorithm>#include <cstring>#include <cstdio>#include <vector>#include <iostream>using namespace std;const int maxn = 1e6 + 10;const int INF =  1e9 + 7;int n,m,d;vector <int> tr[maxn];vector <int> G[maxn];bool tag[maxn] = {0};int f[maxn] = {0};int g[maxn] = {0};bool vis[maxn] = {0};int dp[maxn] = {0};int pre[maxn] = {0};int temp[maxn] = {0};void build (int root) {    vis[root] = 1;    int cnt = tr[root].size();    for (int i = 0;i < cnt; ++ i) {        int u = tr[root][i];        if (!vis[u]) {            pre[u] = root;            G[root].push_back(u);            build (u);        }    }}int dfs1 (int root) {    if (tag[root]) f[root] = 0;    else f[root] = -INF;    int cnt = G[root].size ();    if (cnt == 0) {        return f[root];    }    for (int i = 0;i < cnt; ++ i) {        int u = G[root][i];        dfs1 (u);        f[root] = max (f[root],f[u] + 1);    }    return f[root];}void dfs2 (int root) {    g[root] = g[pre[root]];    if (f[root] + 1 != temp[pre[root]])    g[root] = max (g[root],temp[pre[root]]);    else {        int cnt = G[pre[root]].size();        for (int i = 0;i < cnt; ++ i) {            int u = G[pre[root]][i];            if (u == root) continue;            g[root] = max (g[root],f[u] + 1);        }    }    g[root] = g[root] + 1;    if (tag[root]) g[root] = max (0,g[root]);    int cnt = G[root].size();    for (int i = 0;i < cnt; ++ i) {        int u = G[root][i];        dfs2 (u);    }}int main () {    scanf ("%d%d%d",&n,&m,&d);    for (int i = 1;i <= m; ++ i) {        int x;        scanf ("%d",&x);        tag[x] = 1;    }    for (int i = 1;i < n; ++ i) {        int x,y;        scanf ("%d%d",&x,&y);        tr[x].push_back(y);        tr[y].push_back(x);    }    for (int i = 0;i <= n; ++ i) temp[i] = -INF;    build(1);    dfs1 (1);    for (int i = 1;i <= n; ++ i) {        int cnt = G[i].size();        if (cnt) {            for (int j = 0;j < cnt; ++ j) {                int u = G[i][j];                temp[i] = max (temp[i],f[u] + 1);            }        }    }    g[0] = -INF;    dfs2 (1);    for (int i = 1;i <= n; ++ i) dp[i] = max (f[i],g[i]);    int ans = 0;    for (int i = 1;i <= n; ++ i) {        if (dp[i] >= 0 && dp[i] <= d) {            ans ++;        }    }    printf ("%d\n",ans);    return 0;}

阅读全文

0 0