• Tree


    【原题链接】

    【题意说明】

      指定一棵树上有3个操作:
      (1) 查询a点和b点之间的路径上最长的那条边的长度;
      (2) 区域取反,将a点和b点之间的路径权值都取相反数;
      (3) 单点更新,把某条边的权值变成指定的值。

    【问题分析】
      
    显然对树上进行区域操作,可以用树链剖分来解决。由于有取反操作,所以在线段树的结点上记录区域的最大值和最小值,取反操作时只需要把最小值取反给最大值,最大值取反给最小值。注意取反运算延迟标记的使用(两次取反等于未操作)。

    【参考代码】

    #include<cstdio>

    #include<cstring>

    #include<algorithm>

    using namespace std;

    #define lson k<<1

    #define rson k<<1|1

    #define lt(i) tree[i].left

    #define rt(i) tree[i].right

    #define mx(i) tree[i].max

    #define mn(i) tree[i].min

    #define ct(i) tree[i].cnt

    #define fr(i) edge[i].from

    #define to(i) edge[i].to

    #define nx(i) edge[i].next

    #define vl(i) edge[i].value

    #define dp(i) node[i].deep

    #define tp(i) node[i].top

    #define fa(i) node[i].father

    #define hd(i) node[i].head

    #define sz(i) node[i].size

    #define sn(i) node[i].son

    #define id(i) node[i].id

    const int maxN=10010;

    const int INF=1000000000;

    struct Tree{

             int left, right, max, min, cnt;

    }tree[maxN<<2];

    struct Node{

             int top, head, father, deep, size, son, id;

    }node[maxN];

    struct Edge{

             int from, to, next, value;

    }edge[maxN<<1];

    int n, total;

    void init();

    void work();

    void addedge(int, int, int, int);

    void dfs1(int, int);

    void dfs2(int, int);

    void build(int, int, int);

    void update(int, int, int, int);

    void pushdown(int);

    void pushup(int);

    void update(int, int, int, int, int);

    void change1(int, int);

    void change2(int, int);

    int query(int, int, int);

    void refresh(int, int, int);

    int solve(int, int);

    int main(){

             freopen("poj3237.in", "r", stdin);

             freopen("poj3237.out", "w", stdout);

             int t; scanf("%d", &t);

             while(t--) {init();        work();}

             fclose(stdin);

             fclose(stdout);

             return 0;

    }

    void init(){

             memset(tree, 0, sizeof(tree));

             memset(node, 0, sizeof(node));

             memset(edge, 0, sizeof(edge));

             total=0;

             scanf("%d", &n);

             for(int i=1; i<n; i++){

                       int x, y, c;

                       scanf("%d%d%d", &x, &y, &c);

                       addedge(x, y, c, i);

                       addedge(y, x, c, i+n);

             }

             dfs1(1, 1);

             dfs2(1, 1);

    }

    void addedge(int x, int y, int c, int k){

             fr(k)=x; to(k)=y; vl(k)=c;

             nx(k)=hd(x); hd(x)=k;

    }

    void dfs1(int k, int _dp){

             sz(k)=1; dp(k)=_dp;

             for(int i=hd(k); i; i=nx(i)){

                       if(sz(to(i))) continue;

                       fa(to(i))=k;

                       dfs1(to(i), _dp+1);

                       sz(k)+=sz(to(i));

                       if(sz(sn(k))<sz(to(i))) sn(k)=to(i);

             }

    }

    void dfs2(int k, int _tp){

             id(k)=++total; tp(k)=_tp;

             if(sn(k)) dfs2(sn(k), _tp);

             for(int i=hd(k); i; i=nx(i))

                       if(!id(to(i))) dfs2(to(i), to(i));

    }

    void work(){

             build(1, 1, n);

             for(int i=1; i<n; i++) change1(i, vl(i));

             char ch[10]; int x, y;

             scanf("%s", ch);

             while(ch[0]!='D'){

                       scanf("%d%d", &x, &y);

                       if(ch[0]=='Q')

                                printf("%d ", solve(x, y));

                       else if(ch[0]=='C') change1(x, y);

                       else change2(x, y);

                       scanf("%s", ch);

             }

    }

    void build(int k, int l, int r){

             lt(k)=l; rt(k)=r; mx(k)=-INF; mn(k)=INF;

             if(l==r) return;

             int mid=(l+r)>>1;

             build(lson, l, mid);

             if(mid<r) build(rson, mid+1, r);

    }

    void change1(int k, int w){

             int u=fr(k), v=to(k);

             if(dp(u)<dp(v)) swap(u, v);

             update(1, id(u), id(u), 1, w);

    }

    void change2(int u, int v){

             int f1=tp(u), f2=tp(v);

             while(f1!=f2){

                       if(dp(f1)<dp(f2)){

                                swap(u, v);

                                swap(f1, f2);

                       }

                       update(1, id(f1), id(u), -1, 0);

                       u=fa(f1); f1=tp(u);

             }

             if(u==v) return;

             if(dp(u)<dp(v)) swap(u, v);

             update(1, id(sn(v)), id(u), -1, 0);

    }

    void update(int k, int l, int r, int flag, int w){

             if(l<=lt(k)&&r>=rt(k)){

                       refresh(k, flag, w);

                       return;

             }

             pushdown(k);

             if(l<=rt(lson)) update(lson, l, r, flag, w);

             if(r>=lt(rson)) update(rson, l, r, flag, w);

             pushup(k);

    }

    void refresh(int k, int flag, int w){

             if(flag<0){

                       int t=-mx(k); mx(k)=-mn(k); mn(k)=t;

                       if(!ct(k)) ct(k)=-1; else ct(k)=0;

             }else {mx(k)=mn(k)=w; ct(k)=0;}

    }

    void pushdown(int k){

             if(ct(k)){

                       refresh(lson, ct(k), mx(k));

                       refresh(rson, ct(k), mx(k));

                       ct(k)=0;

             }

    }

    void pushup(int k){

             mx(k)=max(mx(lson), mx(rson));

             mn(k)=min(mn(lson), mn(rson));

    }

    int solve(int u, int v){

             int f1=tp(u), f2=tp(v), ans=-100000;

             while(f1!=f2){

                       if(dp(f1)<dp(f2)) {

                                swap(u, v);

                                swap(f1, f2);

                       }

                       ans=max(ans, query(1, id(f1), id(u)));

                       u=fa(f1); f1=tp(u);

             }

             if(u==v) return ans;

             if(dp(u)<dp(v)) swap(u, v);

             return max(ans, query(1, id(sn(v)), id(u)));

    }

    int query(int k, int l, int r){

             if(l<=lt(k)&&r>=rt(k)) return mx(k);

             pushdown(k);

             int ans=-100000;

             if(l<=rt(lson)) ans=max(ans, query(lson, l, r));

             if(r>=lt(rson)) ans=max(query(rson, l, r), ans);

             return ans;

    }

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  • 原文地址:https://www.cnblogs.com/ahmasoi/p/6678048.html
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