bzoj 4383: [POI2015]Pustynia

Posted 2020-08-27 AωD!

复习了一下线段树优化建图的姿势，在线段树上连边跑拓扑排序

这题竟然卡vector……丧病

 

#include <bits/stdc++.h>
#define N 1810000
using namespace std;
int s, n, m;
int pi[N], vis[N];
int lt[N], nt[N], bi[N], ci[N];
int lp[N], np[N], bp[N];
int tl;
int lb[N], rb[N], lc[N], rc[N];
int out[N], intr[N];
int Q[N], fr, bc;
inline void build(int a, int b, int c)
{
    nt[++ tl] = lt[a]; lt[a] = tl; bi[tl] = b; ci[tl] = c;
    np[tl] = lp[b]; lp[b] = tl; bp[tl] = a; out[a] ++;
    /*
    bi[a].push_back(b);
    bp[b].push_back(a);
    ci[a].push_back(c);
    */
    //printf("%d %d %d\n", a, b, c);
}
int tot;
void dfs_build(int t, int l, int r, int x)
{
    if (l > r) return;
    if (l <= lb[t] && rb[t] <= r) {build(x, t, 0); return;}
    if (l <= rb[lc[t]]) dfs_build(lc[t], l, r, x);
    if (r >= lb[rc[t]]) dfs_build(rc[t], l, r, x);
}
int build_tree(int l, int r)
{
    int t = ++ tot;
    lb[t] = l; rb[t] = r;
    if (l != r)
    {
        lc[t] = build_tree(l, (l + r) / 2); 
        rc[t] = build_tree((l + r) / 2 + 1, r);
        build(t, lc[t], 0);
        build(t, rc[t], 0);
    }
    else intr[l] = t;
    return t;
}

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}

int main()
{
    n = read(); s = read(); m = read();
    build_tree(1, n);
    for (int i = 1; i <= s; ++ i)
    {
        int p, d;
        p = read(); d = read();
        pi[intr[p]] = d;
    }
     
    //return 0;
    for (int i = 1; i <= m; ++ i)
    {
        int l, r, k;
        l = read(); r = read(); k = read();
         
        for (int j = 1, p = l - 1; j <= k + 1; ++ j)
        {
            int a;
            if (j <= k)
            {
                a = read();
                build(intr[a], tot + i, 1);
            }
            else a = r + 1;
            if ((a - 1) - (p + 1) <= 10)
            {
                for (int q = p + 1; q < a; ++ q)
                    build(tot + i, intr[q], 0);
            }
            else 
                dfs_build(1, p + 1, a - 1, tot + i);
            p = a;
        }
    }
    for (int i = 1; i <= m + tot; ++ i)
        if (!out[i]) Q[bc ++] = i;
    
    while (fr != bc)
    {
        int hd = Q[fr ++]; vis[hd] = 1;
        int mx = 1;
        for (int i = lt[hd]; i; i = nt[i])
            mx = max(mx, pi[bi[i]] + ci[i]);
        if (!pi[hd])
        {
            if (mx > 1000000000)
            {
                puts("NIE");
                return 0;
            }
            pi[hd] = mx;
        }
        else if (mx > pi[hd])
        {
            puts("NIE");
            return 0;
        }
        
        for (int i = lp[hd]; i; i = np[i])
        {
            out[bp[i]] --;
            if (!out[bp[i]])
                Q[bc ++] = bp[i];
        }
    }
    for (int i = 1; i <= m + tot; ++ i)
        if (!vis[i]) {puts("NIE"); return 0;}
    puts("TAK");
    for (int i = 1; i <= n; ++ i) printf("%d ", pi[intr[i]]);
}

话说是不是在主席树上也可以干一样的事情呢（手动斜眼