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  • Codeforces 1175E Minimal Segment Cover 倍增

    Minimal Segment Cover

    嗯嗯嗯。。 打倍增的时候两个for写反了查了半天bug。。。

    我们先把没用的线段去掉, 然后对每个线段都找到和它接触的r最大的作为下一个, 然后打倍增就好了。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 5e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    int n, m, c[N];
    int prefix[N];
    int nex[N][20];
    
    struct Line {
        int l, r;
        bool operator < (const Line &rhs) const {
            if(l == rhs.l) return r > rhs.r;
            return l < rhs.l;
        }
    } a[N];
    
    int main() {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d%d", &a[i].l, &a[i].r), a[i].r--;
        sort(a + 1, a + 1 + n);
        int nn = n; n = 1;
        for(int i = 2; i <= nn; i++)
            if(a[i].r > a[n].r) a[++n] = a[i];
        for(int i = 1; i <= n; i++) {
            c[a[i].l]++;
            c[a[i].r + 1]--;
        }
        if(!c[0]) prefix[0] = 1;
        for(int i = 1; i <= 500000; i++) {
            c[i] += c[i - 1];
            if(!c[i]) prefix[i] = 1;
            prefix[i] += prefix[i - 1];
        }
        for(int i = 1, j = 2; i <= n; i++) {
            j = max(i + 1, j);
            while(j <= n && a[j].l <= a[i].r + 1) j++;
            nex[i][0] = j - 1 == i ? 0 : j - 1;
        }
        for(int j = 1; j < 20; j++)
            for(int i = 1; i <= n; i++)
                nex[i][j] = nex[nex[i][j - 1]][j - 1];
        while(m--) {
            int x, y; scanf("%d%d", &x, &y);
            y--;
            int cnt = prefix[y];
            if(x) cnt -= prefix[x - 1];
            if(cnt) {
                puts("-1");
            } else {
                int p = lower_bound(a + 1, a + 1 + n, Line{x + 1, inf}) - a - 1;
                if(a[p].r >= y) {
                    puts("1");
                } else {
                    int ans = 0;
                    for(int i = 19; i >= 0; i--) {
                        if(nex[p][i] && a[nex[p][i]].r < y) {
                            p = nex[p][i], ans += (1 << i);
                        }
                    }
                    printf("%d
    ", ans + 2);
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10982809.html
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