Good Bye 2015 F - New Year and Cleaning

F - New Year and Cleaning

这题简直是丧心病狂折磨王。。

思路：容易想到这样一个转换，把整个矩形一起移动，矩形移出去的时候相当于一行或者一列。

为了优化找到下一个消去的点，我先把原数组扩大两倍，用了st表加二分去找，然后就MLE， 我又换了

线段树TLE，最后不把数组扩大两倍ST表+二分过的。。

每次消去的点都是不变的，所以可以做到线性复杂度。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;

const int N = 5e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;

int k, n, m, X[N], Y[N], Log[N];
char s[N];

struct ST {
    int dp[N][],ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = ; i <= n; i++) dp[i][] = ty * b[i];
        for(int j = ; j <= Log[n]; j++)
            for(int i = ; i+(<<j)- <= n; i++)
                dp[i][j] = max(dp[i][j-], dp[i+(<<(j-))][j-]);
    }
    int query(int x, int y) {
        int k = Log[y - x + ];
        return ty * max(dp[x][k], dp[y-(<<k)+][k]);
    }
} mxx, mnx, mxy, mny;

void walk(int &x, int &y, char c) {
    if(c == 'R') y++;
    else if(c == 'L') y--;
    else if(c == 'U') x--;
    else x++;
}

bool check(int l, int r, int x, int y, int lenr, int lenc) {
    int mxX = mxx.query(l, r) - X[l-];
    int mnX = mnx.query(l, r) - X[l-];
    int mxY = mxy.query(l, r) - Y[l-];
    int mnY = mny.query(l, r) - Y[l-];
    if(x+mxX+lenr > n || x+mnX <  || y+mxY+lenc > m || y+mnY < ) return true;
    return false;
}

int main() {
    for(int i = -(Log[]=-); i < N; i++)
        Log[i] = Log[i - ] + ((i & (i - )) == );

    scanf("%d%d%d%s", &k, &n, &m, s + );
    int x = , y = , L = , R = , D = , U = , p = ;
    for(int i = ; i <= k; i++) {
        walk(x, y, s[i]);
        L = min(L, y); R = max(R, y);
        U = min(U, x); D = max(D, x);
    }
    if(!x && !y) {
        if(R - L +  <= m && D - U +  <= n) {
            puts("-1");
            return ;
        }
    }
    x = , y = ;
    for(int i = ; i <= k; i++) {
        walk(x, y, s[i]);
        X[i] = x; Y[i] = y;
    }

    mxx.build(k, X, );
    mnx.build(k, X, -);
    mxy.build(k, Y, );
    mny.build(k, Y, -);

    x = , y = ;
    int lenr = n, lenc = m;
    LL ans = , t = ;
    while(lenr && lenc) {
        int l = p + , r = min(k, p + k), pos = -;
        while(l <= r) {
            int mid = l + r >> ;
            if(check(p + , mid, x, y, lenr, lenc)) r = mid - , pos = mid;
            else l = mid + ;
        }

        if(pos != -) {
            x += X[pos] - X[p];
            y += Y[pos] - Y[p];
            t = (t + pos - p) % mod;
            p = pos;
            if(s[p] == 'U') {
                lenr--; ans = (ans + 1ll*t*lenc) % mod;
                x = ;
            } else if(s[p] == 'D') {
                lenr--; ans = (ans + 1ll*t*lenc) % mod;
            } else if(s[p] == 'L') {
                lenc--; ans = (ans + 1ll*t*lenr) % mod;
                y = ;
            } else {
                lenc--; ans = (ans + 1ll*t*lenr) % mod;
            }
            if(p == k) p = ;
        } else {
            x += X[k] - X[p];
            y += Y[k] - Y[p];
            t = (t + k - p) % mod;
            p = ;
        }

    }
    printf("%lld\n", ans);
    return ;
}

/*
*/