[NOI2014]魔法森林

链接

loj

思路

a排序，b做动态最小生成树。
把边拆成点就可以了。
uoj98.也许lct复杂度写假了、、越卡常，越慢

代码

#include <bits/stdc++.h>
#define ls c[x][0]
#define rs c[x][1]
using namespace std;
const int N = 2e5 + 7;
int read() {
    int x = 0, f = 1; char s = getchar();
    for (;s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
    for (;s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;        
}
struct edge {
    int x, y, a, b;
    bool operator < (const edge &zz) const {
        return (a^zz.a) ? a < zz.a : b < zz.b;
    }
} G[N];
int f[N], c[N][2], w[N][2], ma[N][2], stak[N], lazy[N], id[N];
bool isroot(int x) {return c[f[x]][0] == x || c[f[x]][1] == x;}
void pushup(int x) {
    ma[x][0] = max(max(ma[ls][0], ma[rs][0]), w[x][0]);
    ma[x][1] = max(max(ma[ls][1], ma[rs][1]), w[x][1]);
    id[x] = (ma[x][1] == w[x][1]) ? x : (ma[ls][1] > ma[rs][1]) ? id[ls] : id[rs];
}
void tag(int x){swap(ls,rs), lazy[x] ^= 1;}
void pushdown(int x) {
    if (lazy[x]) {
        if (ls) tag(ls);
        if (rs) tag(rs);
        lazy[x] ^= 1;
    }
}
void rotate(int x) {
    int y = f[x], z = f[y], k = c[y][1] == x, w = c[x][!k];
    if (isroot(y)) c[z][c[z][1] == y] = x;
    c[x][!k] = y, c[y][k] = w;
    if (w) f[w] = y;
    f[x] = z, f[y] = x;
    pushup(y);
}
void splay(int x) {
    int y = x, z = 0;
    stak[++z] = y;
    while (isroot(y)) stak[++z] = y = f[y];
    while (z) pushdown(stak[z--]);
    while (isroot(x)) {
        y = f[x], z = f[y];
        if (isroot(y)) rotate((c[y][0] == x)^(c[z][0] == y) ? x : y);
        rotate(x);
    }
    pushup(x);
}
void access(int x) {
    for (int y = 0; x;x = f[y = x])
        splay(x), rs = y, pushup(x);
}
void makeroot(int x) {
    access(x), splay(x);
    tag(x);
}
int findroot(int x) {
    access(x), splay(x);
    while(ls) pushdown(x), x = ls;
    return x;
}
void split(int x, int y) {
    makeroot(x), access(y), splay(y);
}
void link(int x, int y) {
    makeroot(x);
    if (findroot(y) != x) f[x] = y;
}
void cut(int x, int y) {
    makeroot(x);
    if (findroot(y) == x && f[x] == y && !rs) {
        f[x] = c[y][0] = 0;
        pushup(y);
    }
}
int main() {
    int n = read(), m = read(), ans = 0x3f3f3f3f;
    for (int i = 1; i <= m; ++i)
        G[i].x = read(), G[i].y = read(), G[i].a = read(), G[i].b = read();
    sort(G + 1, G + 1 + m);
    for (int i = 1; i <= m; ++i) {
        if (G[i].x == G[i].y) continue;
        int x = G[i].x, y = G[i].y;
        if (findroot(x) == findroot(y)) {
            split(x, y);
            if (ma[y][1] > G[i].b) {
                int tmp = id[y];
                cut(G[tmp - n].x, tmp), cut(G[tmp - n].y, tmp);
                w[n + i][0] = G[i].a, w[n + i][1] = G[i].b;
                link(x, n + i), link(n + i, y);
            }
        } else {
            w[n + i][0] = G[i].a, w[n + i][1] = G[i].b;
            link(x, n + i), link(n + i, y);
        }
        if (findroot(1) == findroot(n)) {
            split(1, n);
            ans = min(ans, ma[n][0] + ma[n][1]);
        }
    }
    printf("%d\n", ans == 0x3f3f3f3f ? -1 : ans);
    return 0;
}