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2021CCPC桂林

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题目来自: The 2021 CCPC Guilin Onsite (XXII Open Cup, Grand Prix of EDG)

过了一车的贪心不会…

D - Assumption is All You Need

SOLUTION

贪心，尽量让大的换到前面来让逆序对尽可能多。

CODE
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#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e5 + 10, Log = 20, inf = 0x3f3f3f3f;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n + 1), b(n + 1);
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i <= n; i++) cin >> b[i];
    vector<pii> ans;
    for(int i = 1; i <= n; i++){
        if(a[i] == b[i]) continue;
        if(a[i] < b[i]){
            cout << "-1\n";
            return;
        }
        vector<int> p;
        p.push_back(i);
        for(int j = i + 1; j <= n; j++){
            if(a[j] < a[p.back()] && a[j] >= b[i]){
                p.push_back(j);
            }
            if(a[j] == b[i]) break;
        }
        reverse(p.begin(), p.end());
        for(int j = 0; j + 1 < p.size(); j++){
            ans.emplace_back(p[j], p[j + 1]);
            swap(a[p[j]], a[p[j + 1]]);
        }
        // for(int j = 1; j <= n; j++) cout << a[j] << ' '; cout << '\n';
    }
    cout << ans.size() << '\n';
    for(pii i : ans){
        if(i.first > i.second) swap(i.second, i.first);
        cout << i.first << ' ' << i.second << '\n';
    }
}

int main() {
    int T = 1;
    ios::sync_with_stdio(false);
    cin >> T;
    while (T--) solve();
    return 0;
}

B - A Plus B Problem

SOLUTION

经典线段树区间找最右边的0和9。

虽然经典，但是我vp写的时候bug还是一堆。

CODE
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
const int N=1e6+10;
const int INF=2e9;

int mx[N << 2], mn[N << 2], lz[N << 2];
int ai[N], bi[N], c[N], n;

void apply(int k, int x){
    mx[k] = mn[k] = x;
    lz[k] = x;
}

void pd(int k){
    if(lz[k] == -1) return;
    apply(k << 1, lz[k]);
    apply(k << 1 | 1, lz[k]);
    lz[k] = -1;
}

void pu(int k){
    mx[k] = max(mx[k << 1], mx[k << 1 | 1]);
    mn[k] = min(mn[k << 1], mn[k << 1 | 1]);
}

void build(int k, int l, int r){
    lz[k] = -1;
    if(l == r){
        mx[k] = mn[k] = c[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pu(k);
}

void upd(int k, int l, int r, int ql, int qr, int x){
    if(l > qr || r < ql) return;
    if(l >= ql && r <= qr){
        mx[k] = mn[k] = x;
        lz[k] = x;
        return;
    }
    pd(k);
    int mid = (l + r) >> 1;
    upd(k << 1, l, mid, ql, qr, x);
    upd(k << 1 | 1, mid + 1, r, ql, qr, x);
    pu(k);
}

void upd(int k, int l, int r, int p, int x){
    if(l > p || r < p) return;
    if(l == r){
        mx[k] += x;
        mn[k] += x;
        return;
    }
    pd(k);
    int mid = (l + r) >> 1;
    upd(k << 1, l, mid, p, x);
    upd(k << 1 | 1, mid + 1, r, p, x);
    pu(k);
}

int qry(int k, int l, int r, int p){
    if(l > p || r < p) return -1;
    if(l == r){
        return mx[k];
    }
    pd(k);
    int mid = (l + r) >> 1;
    return max(qry(k << 1, l, mid, p), qry(k << 1 | 1, mid + 1, r, p));
}

int qry0(int k, int l, int r, int ql, int qr){
    if(l > qr || r < ql) return 0;
    if(l == r) return l;
    int mid = (l + r) >> 1;
    pd(k);
    if(l >= ql && r <= qr){
        int res = 0;
        if(mx[k << 1 | 1]) res = max(res, qry0(k << 1 | 1, mid + 1, r, ql, qr));
        else if(mx[k << 1]) res = max(res, qry0(k << 1, l, mid, ql, qr));
        return res;
    }
    int res = 0;
    if(mid < qr && mx[k << 1 | 1]) res = max(qry0(k << 1 | 1, mid + 1, r, ql, qr), res);
    if(mid >= ql && res == 0 && mx[k << 1]) res = max(res, qry0(k << 1, l, mid, ql, qr));
    return res;
}

int qry9(int k, int l, int r, int ql, int qr){
    if(l > qr || r < ql) return 0;
    if(l == r) return l;
    int mid = (l + r) >> 1;
    pd(k);
    if(l >= ql && r <= qr){
        int res = 0;
        if(mn[k << 1 | 1] < 9) res = max(res, qry9(k << 1 | 1, mid + 1, r, ql, qr));
        else if(mn[k << 1] < 9) res = max(res, qry9(k << 1, l, mid, ql, qr));
        return res;
    }
    int res = 0;
    if(mid < qr && mn[k << 1 | 1] < 9) res = max(qry9(k << 1 | 1, mid + 1, r, ql, qr), res);
    if(mid >= ql && res == 0 && mn[k << 1] < 9) res = max(res, qry9(k << 1, l, mid, ql, qr));
    return res;
}

void handle(int ci, int di, int a[]){
    int p = qry(1, 1, n, ci);
    if(a[ci] == di){
        printf("%d 0\n", p);
    }else if(a[ci] > di){ // -
        int x = a[ci] - di;
        if(p >= x){
            upd(1, 1, n, ci, -x);
            printf("%d 2\n", p - x);
        }else{
//            printf("qry0::%d %d\n", 1, ci - 1);
            int p0 = qry0(1, 1, n, 1, ci - 1), cnt = ci - 1 - p0 + 1;
//            printf("!0:: %d\n", p0);
            upd(1, 1, n, p0 + 1, ci - 1, 9);
            p = p + 10 - x;
            upd(1, 1, n, ci, ci, p);
            if(p0){
                upd(1, 1, n, p0, -1);
                cnt++;
            }
            printf("%d %d\n", p, cnt + 1);
        }
    }else{// +
        int x = di - a[ci];
        if(p + x < 10){
            upd(1, 1, n, ci, x);
            printf("%d 2\n", p + x);
        }else{
            int p0 = qry9(1, 1, n, 1, ci - 1), cnt = ci - 1 - p0 + 1;
//            printf("!9:: %d\n", p0);
            upd(1, 1, n, p0 + 1, ci - 1, 0);
            p = p + x - 10;
            upd(1, 1, n, ci, ci, p);
            if(p0){
                upd(1, 1, n, p0, 1);
                cnt++;
            }
            printf("%d %d\n", p, cnt + 1);
        }
    }
    a[ci] = di;
}

void solve(){
    int q;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++){
        scanf("%1d", &ai[i]);
    }
    for(int i = 1; i <= n; i++){
        scanf("%1d", &bi[i]);
    }
    for(int i = n; i >= 1; i--){
        c[i] += ai[i] + bi[i];
        c[i - 1] += c[i] / 10;
        c[i] %= 10;
    }
    build(1, 1, n);
//    for(int i = 1; i <= n; i++) cout << qry(1, 1, n, i);cout << '\n';
    while(q--){
        int ri, ci, di;
        scanf("%d%d%d", &ri, &ci, &di);
        if(ri == 1){
            handle(ci, di, ai);
        }else{
            handle(ci, di, bi);
        }
//        for(int i = 1; i <= n; i++) cout << qry(1, 1, n, i);cout << '\n';
    }
}

int main() {
    int T=1;
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
//    cin>>T;
    //init();
    for(int i=1;i<=T;i++) {
        solve();
    }
}

K - Tax

SOLUTION

看了题解瞬间恍然大悟，顿时感觉自己不太聪明。

在最短路的情况下最多只有$3^{\frac{n}{3}}$种情况，于是爆搜即可。

CODE
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#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e4 + 10, Log = 20, inf = 0x3f3f3f3f;

int d[N], dis[N], v[N], res[N], cnt[N];

struct Edge{
    int v, p;
};

vector<Edge> e[N];

void dfs(int u, int dist){
    res[u] = min(res[u], dist);
    for(Edge i : e[u]){
        int v = i.v, p = i.p;
        if(dis[v] == dis[u] + 1){
            cnt[p]++;
            dfs(v, dist + d[p] * cnt[p]);
            cnt[p]--;
        }
    }
}

void solve() {
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
        cin >> d[i];
    }
    for(int i = 0, u, v, p; i < m; i++){
        cin >> u >> v >> p;
        e[u].push_back({v, p});
        e[v].push_back({u, p});
    }
    queue<int> q;
    q.push(1);
    v[1] = 1;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(Edge i : e[x]){
            if(v[i.v]) continue;
            dis[i.v] = dis[x] + 1;
            v[i.v] = 1;
            q.push(i.v);
        }
    }
    for(int i = 1; i <= n; i++) res[i] = inf;
    dfs(1, 0);
    for(int i = 2; i <= n; i++) cout << res[i] << '\n';
}

int main() {
    int T = 1;
    ios::sync_with_stdio(false);
    // cin >> T;
    while (T--) solve();
    return 0;
}
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