Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
2017-09-07 14:07
537 查看
A
B
题意:n个房子,其中k间房已经住了人,good房间的定义两点,第一点这个房子是空房子,第二点相邻房间至少有一个住了人,k间已经可以住人房间可以任意,问最少和最多有多少间good房子
思路:最少答案要么0要么1, 最多的话可以知道 0 1 0 0 1 0,这样的是在k较少的情况下最多的,这个情况填完后每多一个已经住人的房子,相应的good房就会少一个
C
题意:n个航班,原本n个飞机都是在1到n顺序时间点出发,但由于某些原因所有航班延误了k分钟,也就是说得在k + 1 到 k + n 起飞每架飞机,每架飞机有一个延误权值,每延误一分钟对应会增加相应权值,最后求所有飞机权值和最小,延误后的飞机起飞还有条件,就是起飞时间不能低于原本计划的起飞时间,比如原计划第5分钟飞的飞机在新计划中得5分钟以后起飞(包括第5分钟)
思路:用一个堆去维护前i分钟能起飞的飞机,每次取出权值最大的使他起飞即可
D
题意:这题意很长阿,大致就是这样 n个人去 0 点参加会议,这n个人分布在1 - n中的城市,且一个人仅在一个城市,现在有m个起飞航班和离开航班,所有航班要么从0起飞,要么到达0,现在要求把n个人聚在0点要停留k天,从人到齐的第二天开始算,要准确的停留k天之后才能走,现在问满足n个人能来 0 这个点停留k天并能回去的最小花费。
注意到达那天和离开那天都不能算是停留的天数
思路:用两个数组维护,dpcm[i]代表第i天人全部到齐的最小花费,dpbk[i]代表第i天开始陆续有人离开(并且最终能全部离开)的最小花费,那么很显然答案就是dpcm[i] + dp[i + k + 1]的最小值
之前一直以为是求n个在0人只要待k天即可.....
代码是这样的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 +
4000
7;
const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
struct Node {
LL x;
LL cost;
bool operator < (const Node &w) {
return x < w.x;
}
}p[qq], tmp;
vector<Node> come[qq], back[qq];
int n, m, k;
bool f;
LL sum;
void Judge(int u) {
sort(come[u].begin(), come[u].end());
sort(back[u].begin(), back[u].end());
int i = come[u].size(), j = back[u].size();
if(i == 0 || j == 0) {
f = false;
return;
}
LL minx = 1e18;
i--, j--;
set<LL> mp;
LL l, r;
while(i >= 0) {
while(j >= 0 && come[u][i].x + k + 1 <= back[u][j].x) {
mp.insert(back[u][j].cost);
--j;
}
if(mp.size() > 0) {
minx = min(minx, come[u][i].cost + *(mp.begin()));
}
--i;
}
mp.clear();
if(minx == 1e18) {
f = false;
return;
}
printf("%lld\n", minx);
sum += minx;
}
int main(){
sum = 0;
scanf("%d%d%d", &n, &m, &k);
LL di, fi, ti, ci;
for(int i = 1; i <= m; ++i) {
scanf("%lld%lld%lld%lld", &di, &fi, &ti, &ci);
tmp.x = di, tmp.cost = ci;
if(fi == 0) back[ti].pb(tmp);
else if(ti == 0) come[fi].pb(tmp);
}
f = true;
for(int i = 1; i <= n; ++i) {
Judge(i);
if(!f) break;
}
if(!f) puts("-1");
else printf("%lld\n", sum);
return 0;
}
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <ctime> #include <iostream> #include <algorithm> #include <sstream> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <utility> using namespace std; #define LL long long #define pb push_back #define mk make_pair #define pill pair<int, int> #define mst(a, b) memset(a, b, sizeof a) #define REP(i, x, n) for(int i = x; i <= n; ++i) const int MOD = 1e9 + 7; const int qq = 1e6 + 10; const int INF = 1e9 + 10; int Gcd(int a, int b) { return b == 0 ? a : Gcd(b, a % b); } int main(){ int n; scanf("%d", &n); bool f = false; int a, b; for(int i = 0; i <= n; ++i) { for(int j = i; j <= n; ++j) { if(Gcd(i, j) != 1) continue; if(j == 0) continue; if(i + j == n) { if(!f) { f = true; a = i, b = j; } else { if(a * j < b * i) { a = i, b = j; } } } } } if(a > b) swap(a, b); printf("%d %d\n", a, b); return 0; }
B
题意:n个房子,其中k间房已经住了人,good房间的定义两点,第一点这个房子是空房子,第二点相邻房间至少有一个住了人,k间已经可以住人房间可以任意,问最少和最多有多少间good房子
思路:最少答案要么0要么1, 最多的话可以知道 0 1 0 0 1 0,这样的是在k较少的情况下最多的,这个情况填完后每多一个已经住人的房子,相应的good房就会少一个
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <ctime> #include <iostream> #include <algorithm> #include <sstream> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <utility> using namespace std; #define LL long long #define pb push_back #define mk make_pair #define pill pair<int, int> #define mst(a, b) memset(a, b, sizeof a) #define REP(i, x, n) for(int i = x; i <= n; ++i) const int MOD = 1e9 + 7; const int qq = 1e6 + 10; const int INF = 1e9 + 10; LL n, k; int main(){ scanf("%lld%lld", &n, &k); if(n == k || k == 0 || n == 1) { printf("0 0\n"); return 0; } if(k <= n / 3) { printf("1 %lld\n", k * 2); } else { printf("1 %lld\n", n - k); } return 0; }
C
题意:n个航班,原本n个飞机都是在1到n顺序时间点出发,但由于某些原因所有航班延误了k分钟,也就是说得在k + 1 到 k + n 起飞每架飞机,每架飞机有一个延误权值,每延误一分钟对应会增加相应权值,最后求所有飞机权值和最小,延误后的飞机起飞还有条件,就是起飞时间不能低于原本计划的起飞时间,比如原计划第5分钟飞的飞机在新计划中得5分钟以后起飞(包括第5分钟)
思路:用一个堆去维护前i分钟能起飞的飞机,每次取出权值最大的使他起飞即可
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <ctime> #include <iostream> #include <algorithm> #include <sstream> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <utility> using namespace std; #define LL long long #define pb push_back #define mk make_pair #define pill pair<int, int> #define mst(a, b) memset(a, b, sizeof a) #define REP(i, x, n) for(int i = x; i <= n; ++i) const int MOD = 1e9 + 7; const int qq = 1e6 + 10; const int INF = 1e9 + 10; struct Node { LL t; int id; bool operator < (const Node &w) const { return t < w.t; } }p[qq], tmp; priority_queue<Node> Q; LL n, k; LL ans[qq]; int main(){ scanf("%lld%lld", &n, &k); LL sum = 0; for(int i = 1; i <= n; ++i) { scanf("%lld", &p[i].t); sum += p[i].t; p[i].id = i; } for(int i = 1; i <= k; ++i) { Q.push(p[i]); } LL tot = 0; for(int i = k + 1; i <= n; ++i) { Q.push(p[i]); tmp = Q.top(); Q.pop(); ans[tmp.id] = i; tot += (LL)(i - tmp.id) * tmp.t; } for(int i = n + 1; i <= n + k; ++i) { tmp = Q.top(); Q.pop(); ans[tmp.id] = i; tot += (LL)(i - tmp.id) * tmp.t; } printf("%lld\n", tot); for(int i = 1; i <= n; ++i) { printf("%lld ", ans[i]); } puts(""); return 0; }
D
题意:这题意很长阿,大致就是这样 n个人去 0 点参加会议,这n个人分布在1 - n中的城市,且一个人仅在一个城市,现在有m个起飞航班和离开航班,所有航班要么从0起飞,要么到达0,现在要求把n个人聚在0点要停留k天,从人到齐的第二天开始算,要准确的停留k天之后才能走,现在问满足n个人能来 0 这个点停留k天并能回去的最小花费。
注意到达那天和离开那天都不能算是停留的天数
思路:用两个数组维护,dpcm[i]代表第i天人全部到齐的最小花费,dpbk[i]代表第i天开始陆续有人离开(并且最终能全部离开)的最小花费,那么很显然答案就是dpcm[i] + dp[i + k + 1]的最小值
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <ctime> #include <iostream> #include <algorithm> #include <sstream> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <utility> using namespace std; #define LL long long #define pb push_back #define mk make_pair #define pill pair<int, int> #define mst(a, b) memset(a, b, sizeof a) #define REP(i, x, n) for(int i = x; i <= n; ++i) const int MOD = 1e9 + 7; const int qq = 1e6 + 10; const LL INF = 1e18 + 10; struct Node { int d, x, cost; Node(){} Node(int _d, int _x, int _cost) : d(_d), x(_x), cost(_cost){} bool operator < (const Node &w) const { if(d == w.d) return cost > w.cost; return d > w.d; } }come[qq], back[qq]; int n, m, k; LL dpcm[qq], dpbk[qq]; int c[qq]; bool cmp1(const Node &a, const Node &b) { if(a.d == b.d) return a.cost < b.cost; return a.d < b.d; } bool cmp2(const Node &a, const Node &b) { if(a.d == b.d) return a.cost < b.cost; return a.d > b.d; } int main(){ scanf("%d%d%d", &n, &m, &k); for(int i = 1; i < qq; ++i) { dpcm[i] = dpbk[i] = INF; } int t1, t2; t1 = t2 = 0; int di, fi, ti, ci; for(int i = 0; i < m; ++i) { scanf("%d%d%d%d", &di, &fi, &ti, &ci); if(fi == 0) back[t2++] = Node(di, ti, ci); else come[t1++] = Node(di, fi, ci); } int kind = 0; LL sum = 0; sort(come, come + t1, cmp1); mst(c, -1); for(int i = 0; i < t1; ++i) { if(c[come[i].x] == -1) { sum += come[i].cost; c[come[i].x] = come[i].cost; ++kind; } else if(c[come[i].x] > come[i].cost){ sum += (come[i].cost - c[come[i].x]); c[come[i].x] = come[i].cost; } if(kind == n) dpcm[come[i].d] = min(dpcm[come[i].d], sum); } sort(back, back + t2, cmp2); sum = 0, kind = 0; mst(c, -1); for(int i = 0; i < t2; ++i) { if(c[back[i].x] == -1) { sum += back[i].cost; c[back[i].x] = back[i].cost; ++kind; } else if(c[back[i].x] > back[i].cost) { sum += (back[i].cost - c[back[i].x]); c[back[i].x] = back[i].cost; } if(kind == n) dpbk[back[i].d] = min(dpbk[back[i].d], sum); } for(int i = 2; i < qq - 1; ++i) { dpcm[i] = min(dpcm[i], dpcm[i - 1]); } for(int i = qq - 2; i >= 0; --i) { dpbk[i] = min(dpbk[i], dpbk[i + 1]); } LL ans = INF; for(int i = 1; i < qq - k - 1; ++i) { ans = min(ans, dpcm[i] + dpbk[i + k + 1]); } if(ans == INF) ans = -1; printf("%lld\n", ans); return 0; }
之前一直以为是求n个在0人只要待k天即可.....
代码是这样的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 +
4000
7;
const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
struct Node {
LL x;
LL cost;
bool operator < (const Node &w) {
return x < w.x;
}
}p[qq], tmp;
vector<Node> come[qq], back[qq];
int n, m, k;
bool f;
LL sum;
void Judge(int u) {
sort(come[u].begin(), come[u].end());
sort(back[u].begin(), back[u].end());
int i = come[u].size(), j = back[u].size();
if(i == 0 || j == 0) {
f = false;
return;
}
LL minx = 1e18;
i--, j--;
set<LL> mp;
LL l, r;
while(i >= 0) {
while(j >= 0 && come[u][i].x + k + 1 <= back[u][j].x) {
mp.insert(back[u][j].cost);
--j;
}
if(mp.size() > 0) {
minx = min(minx, come[u][i].cost + *(mp.begin()));
}
--i;
}
mp.clear();
if(minx == 1e18) {
f = false;
return;
}
printf("%lld\n", minx);
sum += minx;
}
int main(){
sum = 0;
scanf("%d%d%d", &n, &m, &k);
LL di, fi, ti, ci;
for(int i = 1; i <= m; ++i) {
scanf("%lld%lld%lld%lld", &di, &fi, &ti, &ci);
tmp.x = di, tmp.cost = ci;
if(fi == 0) back[ti].pb(tmp);
else if(ti == 0) come[fi].pb(tmp);
}
f = true;
for(int i = 1; i <= n; ++i) {
Judge(i);
if(!f) break;
}
if(!f) puts("-1");
else printf("%lld\n", sum);
return 0;
}
相关文章推荐
- 【Codeforces Round #507 (Div. 2, based on Olympiad of Metropolises) A】Palindrome Dance
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) C. Planning(并查集)
- 【Codeforces Round #507 (Div. 2, based on Olympiad of Metropolises) B】Shashlik Cooking
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) C
- Codeforces Round #433 (Div. 1, based on Olympiad of Metropolises) C. Boredom(主席树)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) A. Fraction
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) D. Jury Meeting(双指针模拟)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) ABC
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) A
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)题解
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) B. Maxim Buys an Apartment
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)
- Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)C. Planning