1111 比较麻烦的最短路

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 505;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) struct Node{
int to, nx, di, ti;
}E[N*N*2];
int head[N], tot;
void add(int fr, int to, int di, int ti) {
E[tot].to = to; E[tot].di = di; E[tot].ti = ti; E[tot].nx = head[fr];
head[fr] = tot ++;
}
vector<int> ans1; int Ans1;
vector<int> ans2; int Ans2; struct Hode{
int po, di, ti;
Hode(int a=0, int b=0, int c=0):po(a), di(b), ti(c){}
bool operator < (const Hode & T) const {
if(di != T.di) return di > T.di;
else return ti > T.ti;
}
};
struct Tode{
int po, di, ti;
Tode(int a=0, int b=0, int c=0):po(a), di(b), ti(c){}
bool operator < (const Tode & T) const {
if(ti != T.ti) return ti > T.ti;
else return di > T.di;
}
}; int pre[N], vis[N];
int dis[N], tim[N];
void dfs1(int x, int tag) {
if(x == tag) return;
dfs1(pre[x], tag);
ans1.push_back(x);
}
void dfs2(int x, int tag) {
if(x == tag) return;
dfs2(pre[x], tag);
ans2.push_back(x);
} void dij1(int s, int t) {
priority_queue<Hode> Q;
memset(dis, INF, sizeof(dis));
memset(tim, INF, sizeof(tim));
dis[s] = 0; tim[s] = 0;
memset(vis, 0, sizeof(vis));
Q.push(Hode(s, dis[s], tim[s])); while(!Q.empty()) {
int po = Q.top().po; Q.pop();
if(vis[po]) continue;
vis[po] = 1;
for(int i = head[po]; ~i; i = E[i].nx) {
int to = E[i].to;
if(dis[to] > dis[po] + E[i].di) {
dis[to] = dis[po] + E[i].di;
tim[to] = tim[po] + E[i].ti;
pre[to] = po;
Q.push(Hode(to, dis[to], tim[to]));
}else if(dis[to] == dis[po] + E[i].di && tim[to] > tim[po] + E[i].ti) {
tim[to] = tim[po] + E[i].ti;
pre[to] = po;
Q.push(Hode(to, dis[to], tim[to]));
}
}
}
// printf("Distance = %d: %d", dis[t], s);
ans1.push_back(s); Ans1 = dis[t];
dfs1(t, s);
// printf("\n");
}
void dij2(int s, int t) {
priority_queue<Tode> Q;
memset(dis, INF, sizeof(dis));
memset(tim, INF, sizeof(tim));
dis[s] = 0; tim[s] = 0;
memset(vis, 0, sizeof(vis));
Q.push(Tode(s, dis[s], tim[s])); while(!Q.empty()) {
int po = Q.top().po; Q.pop();
if(vis[po]) continue;
vis[po] = 1;
for(int i = head[po]; ~i; i = E[i].nx) {
int to = E[i].to;
if(tim[to] > tim[po] + E[i].ti) {
dis[to] = dis[po] + 1;
tim[to] = tim[po] + E[i].ti;
pre[to] = po;
Q.push(Tode(to, dis[to], tim[to]));
}else if(tim[to] == tim[po] + E[i].ti && dis[to] > dis[po] + 1) {
dis[to] = dis[po] + 1;
pre[to] = po;
Q.push(Tode(to, dis[to], tim[to]));
}
}
}
ans2.push_back(s); Ans2 = tim[t];
dfs2(t, s);
}
int main() {
int n, m;
while(~scanf("%d %d", &n, &m)) {
ans1.clear(); ans2.clear();
memset(head, -1, sizeof(head));
tot = 0; for(int i = 0; i < m; ++i) {
int a, b, c, d, e; scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);
add(a, b, d, e);
if(!c) add(b, a, d, e);
}
int s,t; scanf("%d %d", &s, &t); dij1(s, t);
dij2(s, t);
// if(s == t) while(1); int fl = 1;
for(int i = 0; i < min(ans1.size(), ans2.size()); ++i) {
if(ans1[i] != ans2[i]) {
fl = 0; break;
}
} if(fl) {
printf("Distance = %d; Time = %d: ", Ans1, Ans2);
for(int i = 0; i < ans1.size(); ++i) {
if(i) printf(" -> ");
printf("%d", ans1[i]);
} printf("\n");
}else {
printf("Distance = %d: ", Ans1);
for(int i = 0; i < ans1.size(); ++i) {
if(i) printf(" -> ");
printf("%d", ans1[i]);
}
printf("\nTime = %d: ", Ans2);
for(int i = 0; i < ans2.size(); ++i) {
if(i) printf(" -> ");
printf("%d", ans2[i]);
}
printf("\n");
}
}
return 0;
}

1112 gcd一下

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 505;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) char s[1005];
map<char, int> mp; int tot;
char table[40];
map<char, int> suc;
vector<int> vc[40]; int gcd(int a, int b) {
if(b == 0) return a;
else return gcd(b, a%b);
}
int main() {
int n;
tot = 0;
for(char i = 'a'; i <= 'z'; ++i) mp[i] = ++tot, table[tot] = i;
for(char i = '0'; i <= '9'; ++i) mp[i] = ++tot, table[tot] = i;
mp['_'] = ++tot; table[tot] = '_'; while(~scanf("%d", &n)) {
suc.clear(); for(int i = 0; i < 40; ++i) vc[i].clear();
scanf("%s", s);
int len = strlen(s);
int tmp = 0;
for(int i = 0; i <= len; ++i) {
if(i && s[i] != s[i-1]) {
vc[mp[s[i-1]]].push_back(tmp);
// printf("%c %d\n", s[i-1], tmp);
tmp = 0;
}
tmp ++;
} for(int i = 1; i <= tot; ++i) {
if(vc[i].size() > 0) {
// printf("%c ", table[i]); for(int j = 0; i < vc[i].size(); ++j) printf("%d ", vc[i][j]); printf("\n");
int tt = vc[i][0];
for(int j = 1; j < vc[i].size(); ++j) {
tt = gcd(tt, vc[i][j]);
} if(tt % n == 0) {
// printf("%c\n",table[i]);
suc[table[i]] = 1;
}
}
} tmp = 0;
for(int i = 0; i < len; ++i) {
if(suc[s[i]] > 0) {
suc[s[i]] = -1;
printf("%c", s[i]);
}
}
printf("\n"); for(int i = 0; i <= len; ++i) {
if(i && s[i] != s[i-1]) {
if(suc[s[i-1]] == -1) {
for(int j = 0; j < tmp/n; ++j) printf("%c", s[i-1]);
}
else for(int j = 0; j < tmp; ++j) printf("%c", s[i-1]);
tmp = 0;
}
tmp ++;
}
printf("\n");
}
return 0;
}

1113

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int A[N]; ll Abs(ll x) {
if(x < 0) return -x;
else return x;
}
int main() {
int n;
while(~scanf("%d", &n)) {
ll all = 0;
for(int i = 0; i < n; ++i) {
scanf("%d", &A[i]);
all += A[i];
}
sort(A, A+n);
ll tmp = 0;
for(int i = 0; i < n/2; ++i) {
tmp += A[i];
}
ll ans = Abs(tmp - (all-tmp));
if(n % 2) {
tmp += A[n/2];
ans = max(ans, Abs(tmp - (all-tmp)) );
} printf("%d %lld\n", n%2, ans); }
return 0;
}

1114 题意不好理解 其实就是并查集

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int f[N];
int find(int x) { return x == f[x]? x : f[x] = find(f[x]); }
void merge(int a, int b) {
if(b == -1 || a == -1) return;
int t1 = find(a); int t2 = find(b); if(t1 != t2) {
if(t1 > t2) f[t1] = t2; else f[t2] = t1;
}
}
int has[N];
int sett[N], area[N];
int cnt[N]; double avg_set[N], avg_area[N];
int ord[N];
int cmp(int x, int y) {
if(avg_area[x] != avg_area[y]) return avg_area[x] > avg_area[y];
else return x < y;
}
int main() {
int n;
while(~scanf("%d", &n)) {
for(int i = 0; i < 1e5; ++i) f[i] = i; for(int i = 0; i < n; ++i) {
int a,b,c; scanf("%d %d %d", &a, &b, &c);
has[a] ++; has[b] ++; has[c] ++;
merge(a, b); merge(a, c);
int d; scanf("%d", &d);
for(int j = 0; j < d; ++j) {
int l; scanf("%d", &l);
has[l] ++;
merge(a, l);
}
scanf("%d %d", &sett[a], &area[a]);
}
// printf("hh\n"); for(int i = 0; i < 1e5; ++i) {
if(has[i]) {
int t1 = find(i);
cnt[t1] ++; avg_set[t1] += sett[i]; avg_area[t1] += area[i];
}
}
int tot = 0;
for(int i = 0; i < 1e5; ++i) {
if(cnt[i]) {
avg_set[i] /= cnt[i]; avg_area[i] /= cnt[i];
ord[tot ++] = i;
}
}
sort(ord, ord + tot, cmp);
printf("%d\n", tot);
for(int i = 0; i < tot; ++i) {
printf("%04d %d %.3f %.3f\n", ord[i], cnt[ord[i]], avg_set[ord[i]], avg_area[ord[i]]);
}
}
return 0;
}

1115

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e3+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int num[N];
int L[N];
int R[N];
int root = 0;
void insert(int x, int nw) {
if(root == -1) {
root = x; return;
}
if(num[x] <= num[nw]) {
if(L[nw]) insert(x, L[nw]);
else L[nw] = x;
}else {
if(R[nw]) insert(x, R[nw]);
else R[nw] = x;
}
}
int cnt[N];
int Deep;
void dfs(int x, int dep) {
Deep = max(Deep, dep);
cnt[dep] ++; if(L[x]) dfs(L[x], dep+1);
if(R[x]) dfs(R[x], dep+1);
}
int main() {
int n;
while(~scanf("%d", &n)) {
root = -1;
for(int i = 0; i < n; ++i) {
scanf("%d", &num[i]);
insert(i, root);
}
Deep = -1;
dfs(root, 0); printf("%d + %d = %d\n", cnt[Deep], cnt[Deep-1], cnt[Deep] + cnt[Deep-1]);
}
return 0;
}

1116

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int Rank[N];
int has[N];
int prime(int x) {
if(x == 1) return 0;
if(x == 2) return 1;
for(int i = 2; i <= sqrt(x); ++i) {
if(x % i == 0)
return 0;
}
return 1;
}
int main() {
int n;
while(~scanf("%d", &n)) {
memset(has, 0, sizeof(has));
for(int i = 0; i < n; ++i) {
int a; scanf("%d", &a);
Rank[a] = i+1;
}
int k; scanf("%d", &k);
while(k--) {
int a; scanf("%d", &a);
if(has[a]) printf("%04d: Checked\n", a);
else if(Rank[a] == 0) printf("%04d: Are you kidding?\n", a);
else if(Rank[a] == 1) printf("%04d: Mystery Award\n", a);
else if(prime(Rank[a])) printf("%04d: Minion\n", a);
else printf("%04d: Chocolate\n", a);
if(Rank[a]) has[a] = 1;
}
}
return 0;
}

1117这题题意也是扯淡啊,,,,

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int a[N];
int main() {
int n;
while(~scanf("%d", &n)) {
for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
sort(a, a+n);
a[n] = INF; int ans = -1;
for(int i = 0; i < n; ++i) {
if(a[i] != a[i+1]) {
ans = max(ans, min(n-i-1, a[i+1]-1) );
}
}
ans = max(ans, min(n, a[0]-1));
printf("%d\n", ans);
}
return 0;
}

1118

include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e4+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int fa[N];
int cnt[N];
int find(int x) { return x == fa[x]? x : fa[x] = find(fa[x]); } int main() {
int n;
while(~scanf("%d", &n)) {
memset(cnt, 0, sizeof(cnt));
for(int i = 1; i <= 1e4; ++i) fa[i] = i; int m = -1;
for(int i = 1; i <= n; ++i) {
int a; scanf("%d", &a);
int pre;
if(a) {
scanf("%d", &pre);
m = max(m, pre);
}
for(int j = 1; j < a; ++j) {
int b; scanf("%d", &b);
m = max(m, b);
int t1 = find(pre); int t2 = find(b);
if(t1 != t2) fa[t2] = t1;
}
}
for(int i = 1; i <= m; ++i) {
cnt[find(i)] ++;
}
int all = 0; int maxx = -1;
for(int i = 1; i <= m; ++i) {
if(cnt[i]) {
all ++;
maxx = max(maxx, cnt[i]);
}
}
printf("%d %d\n", all, m); int k; scanf("%d", &k);
for(int i = 0; i < k; ++i) {
int a; int b; scanf("%d %d", &a, &b);
int t1 = find(a); int t2 = find(b);
if(t1 == t2) printf("Yes\n");
else printf("No\n");
} }
return 0;
}

1119 主要对于先后序的概念需要掌握

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) int pre[N];
int post[N];
int L[N], R[N];
int root; int fl = 1; void solve(int _l, int _r, int l, int r, int rt) {
// printf("%d %d %d %d %d\n", _l,_r,l,r,rt);
if(l > r) return; int ro = pre[_l];
int pos = 0;
for(int i = l; i <= r; ++i) {
if(post[i] == ro) {
pos = i; break;
}
} if(pos == l && pos != r) {
L[rt] = post[pos];
R[rt] = pre[_l+1];
solve(_l+2, _r, l+1, r-1, pre[_l+1]);
}else if(pos == r){
fl = 0;
L[rt] = post[pos];
solve(_l+1, _r, l, r-1, post[pos]);
}else {
// printf("hh %d\n", rt);
L[rt] = post[pos];
int _pos = (pos-l) + _l;
// printf("%d\n", _pos, pre[_pos]);
R[rt] = pre[_pos+1];
solve(_l+1, _pos, l, pos-1, post[pos]);
solve(_pos+2, _r, pos+1, r-1, pre[_pos+1]);
} } int nn = 0;
void dfs(int x) {
if(L[x]) dfs(L[x]);
if(!nn) nn = 1; else printf(" ");
printf("%d", x);
if(R[x]) dfs(R[x]);
}
int main() {
int n;
while(~scanf("%d", &n)) {
memset(L, 0, sizeof(L));
memset(R, 0, sizeof(R));
nn = 0; for(int i = 1; i <= n; ++i) {
scanf("%d", &pre[i]);
}
for(int i = 1; i <= n; ++i) {
scanf("%d", &post[i]);
}
root = pre[1];
solve(2, n, 1, n-1, root); if(fl) printf("Yes\n");
else printf("No\n");
dfs(root);
printf("\n");
}
return 0;
}

1120

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e4+5;
const int INF = 0x3f3f3f3f;
#define MP(x, y) make_pair(x, y) map<int, int> mp;
map<int, int>::iterator it; void solve(int x) {
int tt = 0;
while(x) {
tt += x%10;
x /= 10;
}
mp[tt] ++;
}
int main() {
int n;
while(~scanf("%d", &n)) {
mp.clear(); for(int i = 0; i < n; ++i) {
int a; scanf("%d", &a);
solve(a);
} int cnt = 0;
for(it = mp.begin(); it != mp.end(); ++it) {
if(it->second > 0) {
cnt ++;
}
}
printf("%d\n", cnt);
int fl = 0;
for(it = mp.begin(); it != mp.end(); ++it) {
if(it->second > 0) {
if(!fl) fl = 1; else printf(" ");
printf("%d", it->first);
}
} printf("\n"); }
return 0;
}

pat1111-1120的更多相关文章

  1. CSU 1120 病毒(DP)

    题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1120 解题报告:dp,用一个串去更新另一个串,递推方程是: if(b[i] > a ...

  2. AxureRP8实战手册(基础11-20)

    本文目录 基础11. 设置文本框输入为密码 基础12. 设置打开选择文件窗口 基础13. 限制文本框输入字符位数 基础14. 设置文本框提示文字 基础15. 设置文本框回车触发事件 基础16. 设置元 ...

  3. csuoj 1120: 病毒

    http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1120 1120: 病毒 Time Limit: 3 Sec  Memory Limit: 128 ...

  4. 九度OJ 1120 全排列 -- 实现C++STL中next_permutation()

    题目地址:http://ac.jobdu.com/problem.php?pid=1120 题目描述: 给定一个由不同的小写字母组成的字符串,输出这个字符串的所有全排列. 我们假设对于小写字母有'a' ...

  5. PAT 1120 Friend Numbers

    1120 Friend Numbers (20 分)   Two integers are called "friend numbers" if they share the sa ...

  6. 【Codeforces Round 1120】Technocup 2019 Final Round (Div. 1)

    Codeforces Round 1120 这场比赛做了\(A\).\(C\)两题,排名\(73\). \(A\)题其实过的有点莫名其妙...就是我感觉好像能找到一个反例(现在发现我的算法是对的... ...

  7. 日常英语---十三、MapleStory/Monsters/Level 11-20(邪恶之眼)

    日常英语---十三.MapleStory/Monsters/Level 11-20(邪恶之眼) 一.总结 一句话总结: evil ['ivl] A stronger version of Evil E ...

  8. 1120 Friend Numbers (20 分)

    1120 Friend Numbers (20 分) Two integers are called "friend numbers" if they share the same ...

  9. PAT甲级 1120. Friend Numbers (20)

    1120. Friend Numbers (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Two in ...

  10. PAT 1120 Friend Numbers[简单]

    1120 Friend Numbers (20 分) Two integers are called "friend numbers" if they share the same ...

随机推荐

  1. MySQL系统临时表、用户临时表

    MySQL临时表分为系统使用的临时表和用户使用的临时表. 系统使用的临时表是指MySQL在执行某些SQL语句时需要依赖临时表来完成整个过程.系统使用的临时表的情况可以分为以下几种: *  group ...

  2. MySQL死锁[转]

    案例描述       在定时脚本运行过程中,发现当备份表格的sql语句与删除该表部分数据的sql语句同时运行时,mysql会检测出死锁,并打印出日志. 两个sql语句如下:       (1)inse ...

  3. BZOJ 3720: Gty的妹子树 [树上size分块]

    传送门 题意: 一棵树,询问子树中权值大于$k$的节点个数,修改点权值,插入新点:强制在线 一开始以为询问多少种不同的权值,那道CF的强制在线带修改版,直接吓哭 然后发现看错了这不一道树上分块水题.. ...

  4. POJ 2826 An Easy Problem?![线段]

    An Easy Problem?! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12970   Accepted: 199 ...

  5. EntityFramework 实践 Overview

    使用EntityFramework,是微软出的一个轻量级ORM框架,对于做一些小型的项目非常方便,几乎是零配置,以及对linq的支持,所以非常的易于使用,虽然已经使用EntityFramework很久 ...

  6. adb模拟操作之event

    首语: 我们都知道,adb可以对模拟器和root过的真机进行很多操作,例如:模拟点击,输入,截图,手机和PC,数据互传等.这篇要说的就是adb操作模拟器或者真机的输入输出. 0x01 问题 使用adb ...

  7. 高仿QQ Xplan的H5页面

    概述 这个h5的主要玩法很简单:地球自转的时候会播放背景音乐(比如海浪声),为了找到这个声音是从哪个地球上哪个地方传来的,需要长按下方的按钮,这时地球会自动转动到目标地点,然后镜头拉近,穿过云层,最后 ...

  8. Filezilla Server 出现Error, could not connect to server解决办法

    打开任务管理器:Win+R:services.msc找到Filezilla Server并启动服务

  9. Yii2 Restful Api 401

    采用Yii2 Restful Api方式为APP提供数据,默认你已经做好了所有的编码和配置工作.采用Postman测试接口: 出现这个画面的一个可能原因是:access_token的写法有误,如果你使 ...

  10. CentOS Crontab(定时任务)

    安装crontab: yum install crontabs 说明: service crond start //启动服务 service crond stop //关闭服务 service cro ...