2020 Multi-University Training Contest 4（待补

2020 Multi-University Training Contest 4

1002 Blow up the Enemy

• 思路：贪心 选出造成100伤害所需时间最少的武器数量 计算胜率即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int INF = 0x3f3f3f3f;

int t, n, a, d, cnt, min_;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        min_ = INF, cnt = 0;
        cin >> n;
        for (int i = 1; i <= n; i ++ ){
            cin >> a >> d;
            int tmp = 99 / a * d;
            if (tmp < min_){
                min_ = tmp;
                cnt = 1;
            }
            else if (tmp == min_)
                cnt ++ ;
        }
        cout << 1.0 - (double)1.0 * cnt / (2.0 * n) << "\n";
    }
    return 0;
}

1003 Contest of Rope Pulling

• 思路：裸的01背包 但dp过程中\(\sum \omega_i\)最大值可能达到\(10^6\) 所以可以对数组进行随机排序 缩小背包的大小

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e3 + 10;
const int M = 1e6 + 10;
const ll INF = -1e15;

int t, n, m;
ll sum, sum1, sum2, min_w, ans;
ll dp1[M], dp2[M];

struct node{
    ll w, v;
}a[N], b[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        sum = sum1 = sum2 = ans = 0;
        dp1[0] = dp2[0] = 0;
        cin >> n >> m;
        for (int i = 1; i <= n; i ++ ){
            cin >> a[i].w >> a[i].v;
            sum1 += a[i].w;
        }
        for (int i = 1; i <= m; i ++ ){
            cin >> b[i].w >> b[i].v;
            sum2 += b[i].w;
        }
        random_shuffle(a + 1, a + n + 1);
        random_shuffle(b + 1, b + m + 1);
        min_w = min(sum1, sum2);
        for (int i = 1; i <= min_w; i ++ )
            dp1[i] = dp2[i] = INF;
        for (int i = 1; i <= n; i ++ ){
            sum += a[i].w;
            for (int j = sum; j >= a[i].w; j -- )
                if (dp1[j - a[i].w] != INF)
                    dp1[j] = max(dp1[j], dp1[j - a[i].w] + a[i].v);
        }
        sum = 0;
        for (int i = 1; i <= n; i ++ ){
            sum += b[i].w;
            for (int j = sum; j >= b[i].w; j -- )
                if (dp2[j - b[i].w] != INF)
                    dp2[j] = max(dp2[j], dp2[j - b[i].w] + b[i].v);
        }
        for (int i = 1; i <= min_w; i ++ )
            ans = max(ans, dp1[i] + dp2[i]);
        cout << ans << "\n";
    }
    return 0;
}

1004 Deliver the Cake

• 思路：最短路+拆点 对为MIDDLE的城市建立左手点和右手点 然后对相邻的点连边 如果不同就加上耗费x 注意始点和终点也可能拆点

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

typedef pair<ll, int> p;

const int N = 1e5 + 10;
const int M = 2e5 + 10;
const ll INF = 1e18;

int T, n, m, s, t, tot;
int op[N], head[N << 1];
ll x, ans;
ll dis[N << 1];
string str;

struct Edge{
    int to, nxt;
    ll dist;
}edge[M << 4];

inline void add(int u, int v, ll w){
    edge[ ++ tot ].to = v, edge[tot].dist = w, edge[tot].nxt = head[u], head[u] = tot;
    edge[ ++ tot ].to = u, edge[tot].dist = w, edge[tot].nxt = head[v], head[v] = tot;
}

inline void dij(int st){
    for (int i = 1; i <= 2 * n; i ++ )
        dis[i] = INF;
    dis[st] = 0;
    priority_queue<p, vector<p>, greater<p> > pq;
    pq.push({0, st});
    while (!pq.empty()){
        int u = pq.top().second;
        ll dist = pq.top().first;
        pq.pop();
        if (dis[u] < dist)
            continue;
        for (int i = head[u]; i; i = edge[i].nxt){
            int v = edge[i].to;
            ll dist_ = edge[i].dist;
            if (dis[v] > dis[u] + dist_){
                dis[v] = dis[u] + dist_;
                pq.push({dis[v], v});
            }
        }
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T -- ){
        tot = 0, ans = INF;
        memset(head, -1, sizeof(head));
        cin >> n >> m >> s >> t >> x >> str;
        for (int i = 0; i < str.length(); i ++ ){
            if (str[i] == 'L')
                op[i + 1] = -1;
            else if (str[i] == 'R')
                op[i + 1] = 1;
            else
                op[i + 1] = 0;
        }
        for (int i = 1; i <= m; i ++ ){
            int u, v;
            ll w;
            cin >> u >> v >> w;
            if (op[u] && !op[v])
                swap(u, v);
            if (!op[u]){
                if (!op[v]){
                    add(u, v, w);
                    add(u + n, v + n, w);
                    add(u, v + n, w + x);
                    add(u + n, v, w + x);
                }
                else if (op[v] == 1){
                    add(u, v, w + x);
                    add(u + n, v, w);
                }
                else{
                    add(u, v, w);
                    add(u + n, v, w + x);
                }
            }
            else{
                if (op[u] == op[v])
                    add(u, v, w);
                else
                    add(u, v, w + x);
            }
        }
        dij(s);
        ans = min(ans, min(dis[t], dis[t + n]));
        if (!op[s]){
            dij(s + n);
            ans = min(ans, min(dis[t], dis[t + n]));
        }
        cout << ans << "\n";
    }
    return 0;
}

1005 Equal Sentences

• 思路：水题

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;
const int mod = 1e9 + 7;

int t, n;
int res[N], ans[N];
string s[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        res[1] = ans[1] = 1;
        cin >> n;
        for (int i = 1; i <= n; i ++ )
            cin >> s[i];
        for (int i = 2; i <= n; i ++ ){
            if (s[i] == s[i - 1])
                ans[i] = ans[i - 1];
            else
                ans[i] = (1ll * ans[i - 1] + 1ll * res[i - 1]) % mod;
            res[i] = ans[i - 1];
            // cout << res[i] << " " << ans[i] << "\n";
        }
        cout << ans[n] << "\n";
    }
    return 0;
}

1007 Go Running

• 思路：每个人的起始点为\(x - t\)或\(x + t\) 对这两种情况建边 要用最少的点来覆盖全部的边 问题就转换为了求二分图的最小点覆盖 又因为二分图的最小点覆盖等于二分图的最大匹配 所以用HK算法即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;

int T, n, t, x, na, nb, tot;
int a[N], b[N], aa[N], bb[N], cx[N], cy[N], head[N];
bool vis[N];

struct Edge{
    int to, nxt;
}edge[N << 1];

inline void add(int u, int v){
    edge[ ++ tot ].to = v, edge[tot].nxt = head[u], head[u] = tot;
    edge[ ++ tot ].to = u, edge[tot].nxt = head[v], head[v] = tot;
}

inline bool dfs(int u){
    for (int i = head[u]; i != -1; i = edge[i].nxt){
        int v = edge[i].to;
        if (!vis[v]){
            vis[v] = true;
            if (cy[v] == -1 || dfs(cy[v])){
                cy[v] = u;
                cx[u] = v;
                return true;
            }
        }
    }
    return false;
}

inline int HK(){
    int ans = 0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));
    for (int i = 1; i <= na; i ++ ){
        memset(vis, false, sizeof(vis));
        if (cx[i] == -1 && dfs(i))
            ans ++ ;
    }
    return ans;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T -- ){
        tot = 0;
        memset(head, -1, sizeof(head));
        cin >> n;
        for (int i = 1; i <= n; i ++ ){
            cin >> t >> x;
            a[i] = aa[i] = x - t;
            b[i] = bb[i] = x + t;
        }
        sort(a + 1, a + n + 1), sort(b + 1, b + n + 1);
        na = unique(a + 1, a + n + 1) - a - 1, nb = unique(b + 1, b + n + 1) - b - 1;
        for (int i = 1; i <= n; i ++ ){
            int u = lower_bound(a + 1, a + 1 + na, aa[i]) - a, v = lower_bound(b + 1, b + 1 + nb, bb[i]) - b + na;
            add(u, v);
        }
        cout << HK() << "\n";
    }
    return 0;
}

1011 Kindergarten Physics

• 思路：水题 万有引力常数太小 可忽略不计

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int t, a, b, d, t0;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        cin >> a >> b >> d >> t0;
        cout << d << "\n";
    }
    return 0;
}

1012 Last Problem

• 思路：构造题 在(x,y)处填val 则必须要在周围填val - 1, val - 2, val - 3, val - 4 dfs按左上右下分别是val - 4, val - 2, val - 1, val - 3来构造即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1010;

int n;
int mp[N][N];

inline void dfs(int x, int y, int val){
    if (val <= 0)
        return ;
    if (mp[x + 1][y] != val - 1)
        dfs(x + 1, y, val - 1);
    if (mp[x][y + 1] != val - 2)
        dfs(x, y + 1, val - 2);
    if (mp[x][y - 1] != val - 3)
        dfs(x, y - 1, val - 3);
    if (mp[x - 1][y] != val - 4)
        dfs(x - 1, y, val - 4);
    mp[x][y] = val;
    cout << x << " " << y << " " << val << "\n";
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    dfs(100, 100, n);
    return 0;
}