Codeforces 939F Cutlet dp + 单调队列

Cutlet

写出转移方程式， 发现能用单调队列优化， 写起来比较麻烦。

#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 = 1e5 + 10;
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 dp[107][100007][2];
int n, k, l[107], r[107];
int que[N], be, ed;

int main() {
    memset(dp, 0x3f, sizeof(dp));
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= k; i++) scanf("%d%d", &l[i], &r[i]);
    l[k + 1] = 2 * n;
    dp[0][l[1]][0] = 0;
    dp[0][l[1]][1] = 1;
    dp[0][0][1] = 0;
    dp[0][0][0] = 1;
    for(int i = 1; i <= k; i++) {
        int pt = 0, len = r[i] - l[i];
        be = 1, ed = 0;
        for(int j = 0; j <= n; j++) {
            if(j - len >= 0) dp[i][j][0] = dp[i - 1][j - len][0];
            dp[i][j][1] = dp[i - 1][j][1];
        }
        for(int j = 0; j <= n; j++) {
            while(pt <= j) {
                while(ed >= be && dp[i - 1][pt][0] <= dp[i - 1][que[ed]][0]) ed--;
                que[++ed] = j;
                pt++;
            }
            while(que[be] + len < j) ++be;
            chkmin(dp[i][j][1], dp[i - 1][que[be]][0] + 1);
        }
        be = 1, ed = 0, pt = 0;
        for(int j = 0; j <= n; j++) {
            while(pt <= j) {
                while(ed >= be && dp[i - 1][pt][1] <= dp[i - 1][que[ed]][1]) ed--;
                que[++ed] = j;
                pt++;
            }
            while(que[be] + len < j) ++be;
            chkmin(dp[i][j][0], dp[i - 1][que[be]][1] + 1);
        }
        for(int j = 0; j <= n; j++) {
            chkmin(dp[i][j][0], dp[i][j][1] + 1);
            chkmin(dp[i][j][1], dp[i][j][0] + 1);
        }
        int t = l[i + 1] - r[i];
        if(t) {
            for(int j = n; j >= 0; j--) {
                if(j + t <= n) {
                    chkmin(dp[i][j + t][0], dp[i][j][0]);
                    chkmin(dp[i][j + t][1], dp[i][j][0] + 1);
                }
                dp[i][j][0] = inf;
            }
        }
    }
    if(dp[k][n][0] >= inf) puts("Hungry");
    else {
        puts("Full");
        printf("%d
", min(dp[k][n][0], dp[k][n][1]));
    }
    return 0;
}

/*
*/