2018 CCPC 网络赛
1. spend ai dollars to buy a Power Cube
2. resell a Power Cube and get ai dollars if he has at least one Power Cube
3. do nothing
Obviously, Noswal can own more than one Power Cubes at the same time. After going to the n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.
The first line has an integer n. (1≤n≤105)
The second line has n integers a1,a2,…,an where ai means the trading price (buy or sell) of the Power Cube in the i-th city. (1≤ai≤109)
It is guaranteed that the sum of all n is no more than 5×105.
题意 : 有 n 个城市，在某一个城市可以进行 3 次操作，买一件物品，卖一件物品，或什么也不做，开始的时候有无限的钱，问最终最大的收益是多少？
思路分析 ： 贪心 ， 从第一个城市开始，手头有东西，能卖则卖，卖不掉就买，最后剩下的退掉，但是卖的地方写法很奇特，就是每卖出去一个，我们就压进队列中两个这个元素，第一次弹出队的时候表示反悔了，第二次弹出队的时候表示当时也是将此物品买了，并没有卖出，为什么是这样，纸上写写就知道了
代码示例 ：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn = 1e5+5;
ll n;
struct node
{
ll x, id;
bool operator< (const node &v)const{
if (x == v.x) return id < v.id;
return x > v.x;
}
};
priority_queue<node>que;
int main () {
ll t;
ll x;
cin >> t;
while(t--){
scanf("%lld", &n);
ll ans = 0, cnt = 0;
while(!que.empty()) que.pop();
for(ll i = 1; i <= n; i++){
scanf("%lld", &x);
if (!que.empty() && que.top().x < x) {
if (que.top().id != 2) cnt++;
//printf("===== %lld\n", que.top().id);
ans += x-que.top().x;
que.push({x, 2}); que.push({x, 1});
que.pop();
}
else {
que.push({x, 0});
}
//printf("++++ %lld %lld\n", i, cnt);
}
printf("%lld %lld\n", ans, cnt*2);
}
return 0;
}
Problem Description
There are N vertices connected by N?1 edges, each edge has its own length.
The set { 1,2,3,…,N } contains a total of N! unique permutations, let’s say the i-th permutation is Pi and Pi,j is its j-th number.
For the i-th permutation, it can be a traverse sequence of the tree with N vertices, which means we can go from the Pi,1-th vertex to the Pi,2-th vertex by the shortest path, then go to the Pi,3-th vertex ( also by the shortest path ) , and so on. Finally we’ll reach the Pi,N-th vertex, let’s define the total distance of this route as D(Pi) , so please calculate the sum of D(Pi) for all N! permutations.
Input
There are 10 test cases at most.
The first line of each test case contains one integer N ( 1≤N≤105 ) .
For the next N?1 lines, each line contains three integer X, Y and L, which means there is an edge between X-th vertex and Y-th of length L ( 1≤X,Y≤N,1≤L≤109 ) .
Output
For each test case, print the answer module 109+7 in one line.
Sample Input
3
1 2 1
2 3 1
3
1 2 1
1 3 2
Sample Output
16
24
题意 ： 初始状态下给你一棵树，同时给你边权上的值，此时整棵树的值为所有边权的累加，然后让你求 n! 下树的边权的累加的和。
思路分析 ：加上 n! 这个限制条件，题目差不多就转换成初始状态下任意点对的距离和，很像，但不完全是
在这个问题中可以发现的任意一个点对在 n! 中出现的次数是 2*(n-1)! 的，任意打一个表就可以看出，然后在某一个图下，我是计算的时候是枚举任意一条边的，出现的次数等于边左边的点乘以边右边的点，此时在乘以 2*(n-1)! 即可
代码示例 ：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn = 1e5+5;
const ll mod = 1e9+7;
ll n;
ll pp[maxn];
void init() {
pp[0] = 1;
for(ll i = 1; i <= 100000; i++){
pp[i] = pp[i-1]*i%mod;
}
}
struct node
{
ll to, cost;
};
vector<node>ve[maxn];
ll size[maxn];
ll ans;
void dfs(ll x, ll fa, ll w){
for(ll i = 0; i < ve[x].size(); i++){
ll to = ve[x][i].to;
if (to == fa) continue;
dfs(to, x, ve[x][i].cost);
size[x] += size[to];
}
size[x] += 1;
ans += size[x]*(n-size[x])%mod*w%mod*2%mod*pp[n-1]%mod;
ans %= mod;
}
int main () {
ll u, v, w;
init();
while(~scanf("%lld", &n)){
for(int i = 1; i <= 100000; i++) ve[i].clear();
for(ll i = 1; i < n; i++){
scanf("%lld%lld%lld", &u, &v, &w);
ve[u].push_back({v, w});
ve[v].push_back({u, w});
}
memset(size, 0, sizeof(size));
ans = 0;
dfs(1, 0, 0);
printf("%lld\n", ans);
}
return 0;
}
YJJ's Salesman
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1369 Accepted Submission(s): 483
Problem Description
YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109). YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109), he will only forward to (x+1,y), (x,y+1) or (x+1,y+1).
On the rectangle map from (0,0) to (109,109), there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109), only the one who was from (xk?1,yk?1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.
Input
The first line of the input contains an integer T (1≤T≤10),which is the number of test cases.
In each case, the first line of the input contains an integer N (1≤N≤105).The following N lines, the k-th line contains 3 integers, xk,yk,vk (0≤vk≤103), which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.
Output
The maximum of dollars YJJ can get.
Sample Input
1
3
1 1 1
1 2 2
3 3 1
Sample Output
3
题意 ： 在一张 10^9 * 10^9 的图中有 1e5 个点，规定只能向右走，向下走，向右下走，只有当往右下走的时候进入格点才会获得此中的分数，问最终最大的得分是多少？
思路分析 ： 首先离散一下，将图变成 1e5*1e5 的图，我们一列一列的去维护，对于某一列中的某个点，如果想要得分，一定是他前一列中他所在位置上方的点转移下来的，就按照这个思路写就可以了
代码示例：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn = 1e5+5;
#define lson k<<1
#define rson k<<1|1
ll n;
ll x[maxn], y[maxn];
struct pp
{
ll u, v, w;
bool operator< (const pp &q)const{
if (u == q.u) return v < q.v;
return u < q.u;
}
}arr[maxn];
struct node
{
ll l, r;
ll mm;
}t[maxn<<2];
void build(ll l, ll r, ll k){
t[k].l = l, t[k].r = r; t[k].mm = 0;
if (l == r) return;
ll mid = (l+r)>>1;
build(l, mid, lson);
build(mid+1, r, rson);
}
ll query(ll l, ll r, ll k){
if (l <= t[k].l && t[k].r <= r){
return t[k].mm;
}
ll mid = (t[k].l+t[k].r)>>1;
ll res = 0;
if (l <= mid) res = max(res, query(l, r, lson));
if (r > mid) res = max(res, query(l, r, rson));
return res;
}
void update(ll pos, ll val, ll k){
if (t[k].l == t[k].r) {
t[k].mm = max(t[k].mm, val);
return;
}
ll mid = (t[k].l+t[k].r)>>1;
if (pos <= mid) update(pos, val, lson);
else update(pos, val, rson);
t[k].mm = max(t[lson].mm, t[rson].mm);
}
int main () {
ll T;
cin >> T;
while(T--){
scanf("%lld", &n);
ll w;
for(ll i = 1; i <= n; i++){
scanf("%lld%lld%lld", &x[i], &y[i], &w);
arr[i] = {x[i], y[i], w};
}
sort(x+1, x+1+n); sort(y+1, y+1+n);
ll ss = unique(x+1, x+1+n)-x;
ll ss2 = unique(y+1, y+1+n)-y;
for(ll i = 1; i <= n; i++){
arr[i].u = lower_bound(x+1, x+ss, arr[i].u)-x;
arr[i].v = lower_bound(y+1, y+ss2, arr[i].v)-y;
}
sort(arr+1, arr+1+n);
// for(ll i = 1; i <= n; i++) {
// prllf("+++ %d %d %d\n", arr[i].u, arr[i].v, arr[i].w);
// }
build(1, n, 1);
for(ll i = 1; i <= n; i++){
ll p = i+1;
while(arr[p].u == arr[i].u) p++;
for(ll j = p-1; j >= i; j--){
ll f = query(1, arr[j].v-1, 1);
update(arr[j].v, arr[j].w+f, 1);
}
i = p-1;
}
printf("%lld\n", t[1].mm);
}
return 0;
}
