UOJ 274 温暖会指引我们前进 - LCT

Solution

更新掉路径上温暖度最小的边就可以了~

Code

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define ll long long
 #define rd read()
 using namespace std;

 const int N = 4e5 + ;
 const int inf = ~0U >> ;

 int n, m, fa[N];

 struct edge {
     int u, v, t, w;
 }e[N];

 int read() {
     int X = , p = ; char c = getchar();
     for (; c > '' || c < ''; c = getchar())
         if (c == '-') p = -;
     for (; c >= '' && c <= ''; c = getchar())
         X = X *  + c - '';
     return X * p;
 }

 int anc(int x) {
     return fa[x] == x ? x : fa[x] = anc(fa[x]);
 }

 namespace LCT {
     int f[N], ch[N][], val[N], mn[N], tun[N];
     ll sum[N];
 #define lc(x) ch[x][0]
 #define rc(x) ch[x][1]

     int isroot(int x) {
         return lc(f[x]) != x && rc(f[x]) != x;
     }

     int get(int x) {
         return rc(f[x]) == x;
     }

     void up(int x) {
         mn[x] = val[x];
         sum[x] = e[val[x]].t;
         if (e[mn[lc(x)]].w < e[mn[x]].w) mn[x] = mn[lc(x)];
         if (e[mn[rc(x)]].w < e[mn[x]].w) mn[x] = mn[rc(x)];
         if (lc(x)) sum[x] += sum[lc(x)];
         if (rc(x)) sum[x] += sum[rc(x)];
     }

     void rev(int x) {
         swap(lc(x), rc(x));
         tun[x] ^= ;
     }

     void pushdown(int x) {
         if (tun[x]) {
             if (lc(x)) rev(lc(x));
             if (rc(x)) rev(rc(x));
             tun[x] = ;
         }
     }

 int st[N], tp;

     void pd(int x) {
         while (!isroot(x)) {
             st[++tp] = x;
             x = f[x];
         }
         pushdown(x);
         while (tp) pushdown(st[tp--]);
     }

     void rotate(int x) {
         int old = f[x], oldf = f[old], son = ch[x][get(x) ^ ];
         if (!isroot(old)) ch[oldf][get(old)] = x;
         ch[x][get(x) ^ ] = old;
         ch[old][get(x)] = son;
         f[old] = x; f[x] = oldf; f[son] = old;
         up(old); up(x);
     }

     void splay(int x) {
         pd(x);
         for (; !isroot(x); rotate(x))
             if (!isroot(f[x]))
                 rotate(get(f[x]) == get(x) ? f[x] : x);
     }

     void access(int x) {
         for (int y = ; x; y = x, x = f[x])
             splay(x), ch[x][] = y, up(x);
     }

     int findr(int x) {
         access(x); splay(x);
         while (lc(x)) pushdown(x), x = lc(x);
         return x;
     }

     void mroot(int x) {
         access(x); splay(x); rev(x);
     }

     void split(int x, int y) {
         mroot(x); access(y); splay(y);
     }

     void link(int x, int y) {
         mroot(x);
         f[x] = y;
     }

     void cut(int x, int y) {
         split(x, y);
         f[x] = ch[y][] = ;
     }
 }
 using namespace LCT;

 int main()
 {
     e[].w = inf;
     n = rd; m = rd;
     for (int i = ; i <= n; ++i)
         fa[i] = i;
     for (int i = ; i <= m; ++i)
         val[i + n] = i;
     char op[];
     for (int i = ; i <= m; ++i) {
         scanf("%s", op);
         if (op[] == 'f') {
             int id = rd + ;
             e[id].u = rd + ;
             e[id].v = rd + ;
             e[id].w = rd;
             e[id].t = rd;
             int x = e[id].u, y = e[id].v;
             if (anc(x) != anc(y)) {
                 link(id + n, x);
                 link(id + n, y);
                 x = anc(x); y = anc(y);
                 fa[x] = y;
             }
             else {
                 split(x, y);
                 int t = mn[y];
                 if (e[id].w < e[t].w)
                     continue;
                 cut(t + n, e[t].u);
                 cut(t + n, e[t].v);
                 link(id + n, e[id].u);
                 link(id + n, e[id].v);
             }
         }
         else if(op[] == 'm') {
             int x = rd + , y = rd + ;
             if (anc(x) != anc(y)) {
                 puts("-1"); continue;
             }
             if (x == y) {puts(""); continue;}
             split(x, y);
             printf("%lld\n", sum[y]);
         }
         else {
             int id = rd + ;
             e[id].t = rd;
             mroot(id + n);
         }
     }
 }