Codeforces 542A Place Your Ad Here

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把没用的第一类区间去掉之后， 排序， 然后枚举第二类区间， 在上面死命二分就好了。

#include<bits/stdc++.h>
#define LL 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 ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);

bool chkmax(LL& a, LL b) {
    return a < b ? a = b, true : false;
}
bool chkmax(int& a, int b) {
    return a < b ? a = b, true : false;
}

int Log[N];
struct ST {
    PII dp[N][20];
    void build(int n, PII b[]) {
        for(int i = 1; i <= n; i++) dp[i][0] = b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    PII query(int x, int y) {
        int k = Log[y - x + 1];
        return max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
};

int n, m;
PII len[N];
PII gg = mk(1, 1);

pair<PII, int> a[N];
pair<PII, int> b[N];
ST rmq;

bool cmp(const pair<PII, int>& a, const pair<PII, int>& b) {
    if(a.fi.fi == b.fi.fi) return a.fi.se > b.fi.se;
    return a.fi.fi < b.fi.fi;
}
int main() {
    for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d%d", &a[i].fi.fi, &a[i].fi.se), a[i].se = i;
    for(int i = 1; i <= m; i++) scanf("%d%d%d", &b[i].fi.fi, &b[i].fi.se, &b[i].se);
    sort(a + 1, a + 1 + n, cmp);
    int up = n; n = 1;
    for(int i = 1; i <= up; i++)
        if(a[i].fi.se > a[n].fi.se) a[++n] = a[i];
//    for(int i = 1; i <= n; i++) printf("%d %d %d ^^\n", a[i].se, a[i].fi.fi, a[i].fi.se);
    for(int i = 1; i <= n; i++) len[i].fi = a[i].fi.se - a[i].fi.fi, len[i].se = a[i].se;
    rmq.build(n, len);
    LL ans = 0;
    for(int i = 1; i <= m; i++) {
        int L = b[i].fi.fi, R = b[i].fi.se, c = b[i].se, len = 0, who = 1;
        int low = 1, high = n, p = -1;
        while(low <= high) {
            int mid = low + high >> 1;
            if(a[mid].fi.fi <= L) p = mid, low = mid + 1;
            else high = mid - 1;
        }
        if(~p && a[p].fi.se > L && chkmax(len, min(R, a[p].fi.se) - L)) who = a[p].se;
        low = 1, high = n, p = -1;
        while(low <= high) {
            int mid = low + high >> 1;
            if(a[mid].fi.se >= R) p = mid, high = mid - 1;
            else low = mid + 1;
        }
        if(~p && a[p].fi.fi < R && chkmax(len, R - max(L, a[p].fi.fi))) who = a[p].se;
        int Lb = -1, Rb = -1;
        low = 1, high = n;
        while(low <= high) {
            int mid = low + high >> 1;
            if(a[mid].fi.fi >= L) Lb = mid, high = mid - 1;
            else low = mid + 1;
        }
        low = 1, high = n;
        while(low <= high) {
            int mid = low + high >> 1;
            if(a[mid].fi.se <= R) Rb = mid, low = mid + 1;
            else high = mid - 1;
        }
        if(~Lb && ~Rb && Lb <= Rb) {
            PII t = rmq.query(Lb, Rb);
            if(chkmax(len, t.fi)) who = t.se;
        }
        if(chkmax(ans, 1LL * c * len)) {
            gg.fi = who;
            gg.se = i;
        }
    }
    if(ans) {
        printf("%lld\n", ans);
        printf("%d %d\n", gg.fi, gg.se);
    } else {
        puts("0");
    }
    return 0;
}

/*
*/