2019 Multi-University Training Contest 9
2019-08-20 / Luo Jinrong   

进度

A B C D E F G H I J K
√ √ ∅

1002. Rikka with Cake

题意

一个$n\times m$的蛋糕，有$k$条射线，求这个蛋糕被分成几块。

做法

离散化+树状数组。求这$k$条射线的交点数，答案即为交点数+1。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define per(i,a,n) for (int i=a;i<n;i++)
#define rep(i,a,n) for (int i=n-1;i>=a;i--)

const int MAXN(5e5+7);
int T;
ll n, m, x, y;
ll posh, poss, pos, K, cntx, cnty;
char Dir[] = {'L', 'R', 'U', 'D'}, Dd;
struct nd{
    ll x, y;
    ll D;
}hln[MAXN], sln[MAXN];
ll hashx[MAXN], hashy[MAXN];
ll outt;

inline ll Lowbit(ll k) { return (k&-k); }

ll sizze;
struct Ln {
    ll c[MAXN];
    inline void init() {
        per(i, 0, sizze+1) c[i]=0;
    }
    inline void update(ll _pos) {
        while (_pos <= sizze) { //注意这里
            c[_pos] += 1ll;
            _pos += Lowbit(_pos);
        }
    }
    inline ll sum(ll _pos) {
        ll s = 0;
        while (_pos > 0) {
            s += c[_pos];
            _pos -= Lowbit(_pos);
        }
        return s;
    }
}LT, RT;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> T;
    while(T--) {
        outt = posh = poss = outt = cntx = cnty = 1;
        cin >> n >> m >> K;
        per(i, 1, K+1) {
            cin >> x >> y >> Dd;
            hashx[cntx++] = x;
            hashy[cnty++] = y;
            per(j, 0, 4) {
                if(Dd==Dir[j]) {
                    if(j<2) {
                        hln[posh].D = j;
                        hln[posh].x = x;
                        hln[posh++].y = y;
                    } else {
                        sln[poss].D = j;
                        sln[poss].x = x;
                        sln[poss++].y = y;
                    }
                    break;
                }
            }
        }

        //离散化
        sort(hashx+1, hashx+cntx);
        sort(hashy+1, hashy+cnty);
        cntx = unique(hashx+1, hashx+cntx)-hashx;
        cnty = unique(hashy+1, hashy+cnty)-hashy;
        per(i, 1, poss) {
            sln[i].x = lower_bound(hashx+1, hashx+cntx, sln[i].x)-hashx;
            sln[i].y = lower_bound(hashy+1, hashy+cnty, sln[i].y)-hashy;
        }
        per(i, 1, posh) {
            hln[i].x = lower_bound(hashx+1, hashx+cntx, hln[i].x)-hashx;
            hln[i].y = lower_bound(hashy+1, hashy+cnty, hln[i].y)-hashy;
        }
        //离散化完成

        sizze = cntx;
        pos = poss;
        sort(sln+1, sln+poss, [](const nd& a, const nd& b) {return a.D==b.D?a.y>b.y:a.D<b.D;});
        //向上线处理
        LT.init();
        RT.init();
            //对横向线从上到下排序
        int tmph=1;
        ll lx=0, rx=cntx, tmpy=cnty;
        sort(hln+1, hln+posh, [](const nd& a, const nd& b){return a.y>b.y;});
        per(i, 1, poss) {
            if(sln[i].D==3) {
                pos = i;
                break;
            }
            while(tmph<posh && hln[tmph].y>=sln[i].y) {
                //判断是否还在上一个一行
                if(hln[tmph].y==tmpy) {
                    if(!hln[tmph].D) {
                        lx = max(lx, hln[tmph].x);
                    } else {
                        rx = min(rx, hln[tmph].x);
                    }
                } else {
                    if(lx>=rx) {
                        lx = 0;
                        rx = 1;
                    }
                    //处理树状数组
                    if(lx) LT.update(lx);
                    if(rx<cntx) RT.update(rx);
                    tmpy=hln[tmph].y, lx=0, rx=cntx;
                    if(!hln[tmph].D) {
                        lx = hln[tmph].x;
                    } else {
                        rx = hln[tmph].x;
                    }
                }
                tmph++;
            }
            if(tmpy>=sln[i].y) {
                if(lx>=rx) {
                    lx = 0;
                    rx = 1;
                }
                //处理树状数组
                if(lx) LT.update(lx);
                if(rx<cntx) RT.update(rx);
                if(tmph<posh) {
                    tmpy=hln[tmph].y, lx=0, rx=cntx;
                    if (!hln[tmph].D) {
                        lx = hln[tmph].x;
                    } else {
                        rx = hln[tmph].x;
                    }
                } else {
                    tmpy=0;
                }
            }
            //判断穿过多少条线
            outt += (LT.sum(sizze)-LT.sum(sln[i].x-1));
            outt += RT.sum(sln[i].x);
        }
        //向下线处理
        LT.init();
        RT.init();
        tmph=1, lx=0, rx=cntx;
        tmpy=0;
        sort(hln+1, hln+posh, [](const nd& a, const nd& b){return a.y<b.y;});
        rep(i, pos, poss) {
            while(tmph<posh && hln[tmph].y<=sln[i].y) {
                //判断是否还在上一个一行
                if(hln[tmph].y==tmpy) {
                    if(!hln[tmph].D) {
                        lx = max(lx, hln[tmph].x);
                    } else {
                        rx = min(rx, hln[tmph].x);
                    }
                } else {
                    if(lx>=rx) {
                        lx = 0;
                        rx = 1;
                    }
                    //处理树状数组
                    if(lx) LT.update(lx);
                    if(rx<cntx) RT.update(rx);
                    tmpy=hln[tmph].y, lx=0, rx=cntx;
                    if(!hln[tmph].D) {
                        lx = hln[tmph].x;
                    } else {
                        rx = hln[tmph].x;
                    }
                }
                tmph++;
            }
            if(tmpy<=sln[i].y) {
                if(lx>=rx) {
                    lx = 0;
                    rx = 1;
                }
                //处理树状数组
                if(lx) LT.update(lx);
                if(rx<cntx) RT.update(rx);
                if(tmph<posh) {
                    tmpy=hln[tmph].y, lx=0, rx=cntx;
                    if (!hln[tmph].D) {
                        lx = hln[tmph].x;
                    } else {
                        rx = hln[tmph].x;
                    }
                } else {
                    tmpy = cnty;
                }
            }
            //判断穿过多少条线
            outt += (LT.sum(sizze)-LT.sum(sln[i].x-1));
            outt += RT.sum(sln[i].x);
        }
        cout << outt << endl;
    }

    return 0;
}

1005. Rikka with Game

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#pragma warning(disable:4996)
#include<bits/stdc++.h>
#define upf(a,b,c) for(int a=b;a<=c;++a)
#define drf(a,b,c) for(int a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define Tcase(n) for(int _=1;_<=n;++_)
#define ll long long
#define ull unsigned long long
#define ShaShiBuGan bool woyebuzhidaogaiganmasuoyishenmedoubugan=false;
const ll mod(1e9);
const int maxn(105);
const ll INF(0x3fffffffff);
const double sfive(sqrt(5));
using namespace std;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
int T;
int n, m, x, y, k;
char s[maxn * 3];
ll dp[maxn];
ll mp[maxn][maxn], num[maxn], marked[maxn];

int main() {
#ifdef ACM_LOCAL
    freopen("./ACM.in", "r", stdin);
    freopen("./ACM.out", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    Tcase(T) {
        cin >> s;
        int length = strlen(s);
        upf(i, 0, length - 1) {
            if (s[i] != 'y') {
                if (s[i] == 'z')
                    s[i] = 'b';
                break;
            }
        }
        cout << s << endl;
    }
}

return 0;

本文链接：
https://luojinrong.github.io/2019/08/20/2019-Multi-University-Training-Contest-9/