Contest3898 - 计科23级算法设计与分析平时作业-01
题目链接
A.DNA Sorting
题面
思路
题目意思就是说，如果一个字符串中前面的字符比后面的字符大，那么它的无序度就+1，根据这个给一组字符串从最有序到最无序依次输出。那么明白题目意思之后直接模拟即可。
示例代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'
const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector<pair<int, string>> arr(m);
fer(i, 0, m) cin >> arr[i].second;
for(auto &p : arr){
string s = p.second;
fer(i, 0, s.size()){
fer(j, i + 1, s.size()) p.first += (s[i] > s[j]);
}
}
sort(all(arr));
for(auto p : arr) cout << p.second << '\n';
return 0;
}
B.Integer Inquiry
题面
思路
其实就是求多个数的高精度加法
示例代码
ans = 0
while True:
n = int(input())
if n == 0:
break
ans += n
print(ans)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'
const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;
string add(string s1, string s2) {
string res;
vector<int> a(110), b(110), c(110);
int n1 = s1.size(), n2 = s2.size();
for(int i = 0, j = n1 - 1; i < n1; i++, j--) a[i] = s1[j] - '0';
for(int i = 0, j = n2 - 1; i < n2; i++, j--) b[i] = s2[j] - '0';
fer(i, 0, max(n1, n2)){
c[i] += a[i] + b[i];
c[i + 1] += c[i] / 10;
c[i] %= 10;
}
int len = max(n1, n2) + (c[max(n1, n2)] != 0);
ferd(i, len - 1, 0) res += c[i] + '0';
return res;
}
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
string s;
string ans = "";
while(cin >> s, s != "0"){
ans = add(ans, s);
}
cout << ans << '\n';
return 0;
}
C.Word Amalgamation
题面
思路
题意是说先输入一组词典，词典中都是正确的单词，然后输入一组乱序的词，这些乱序的词可以经过重新将字母排序得到词典中的单词（或者词典中不存在）。如果存在就依次输出乱序单词对应的正确单词，不存在则另输出。
思路：我们只需要比较乱序单词和词典中的单词中的各字母的频次即可。
示例代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'
const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
string s;
map<string, map<int, int>> mp1;
map<string, map<int, int>> mp2;
while(cin >> s, s != "XXXXXX"){
for(auto c : s) mp1[s][c - 'a']++;
}
vector<string> arr;
while(cin >> s, s != "XXXXXX"){
arr.push_back(s);
for(auto c : s) mp2[s][c - 'a']++;
}
for(auto x : arr){
bool ok = false;
for(auto [y, mp] : mp1){
if(mp1[y] == mp2[x]){
ok = true;
cout << y << '\n';
}
}
if(!ok) cout << "NOT A VALID WORD\n";
cout << "******\n";
}
return 0;
}
D.Sorting It All Out
题面
思路
示例代码
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAXN=2500;
int n,m,cas,id;
struct edge
{
int v,nx;
}set[MAXN];
int head[30],d[30],ok[30],write[30];
queue<int> Q;
void Addedge(int u,int v)
{
id++;set[id].v=v;set[id].nx=head[u];
head[u]=id;
}
int bfs()
{
cas=0;
while(!Q.empty())Q.pop();
int out=1,t,u;t=0;
for(int i=1;i<=n;i++)
if(!ok[i])
{
t++;Q.push(i);
}
if(t>1)out=0;
while(!Q.empty())
{
u=Q.front();Q.pop();t=0;write[++cas]=u;
for(int k=head[u];k>0;k=set[k].nx)
{
ok[set[k].v]--;
if(!ok[set[k].v])
{
t++;Q.push(set[k].v);
}
}
if(t>1)out=0;
}
for(int i=1;i<=n;i++)if(ok[i]>0){out=-1;break;}
return out;
}
char print()
{
for(int i=1;i<=cas;i++)printf("%c",(char)(write[i]+64));
return '.';
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(write,0,sizeof(write));
memset(d,0,sizeof(d));
memset(head,-1,sizeof(head));
id=0;
int now=0,loca;
char a,b,c;
if(n==0 && m==0)break;
for(int i=1;i<=m;i++)
{
cin>>a>>b>>c;
if(b=='<')
{
Addedge(a-64,c-64);d[c-64]++;
}
else
{
Addedge(c-64,a-64);d[a-64]++;
}
if(now==0)
{
for(int j=1;j<=n;j++)ok[j]=d[j];
now=bfs();loca=i;
}
}
if(now==0)puts("Sorted sequence cannot be determined.");
else if(now==1)printf("Sorted sequence determined after %d relations: ",loca),printf("%c\n",print());
else printf("Inconsistency found after %d relations.\n",loca);
}
return 0;
}
E.Cable TV Network
题面
示例代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'
const int MOD = 1e9 + 7;
const int inf = 1e9;
const int N = 105, M = 10005;
int n, m, T, head[N], ver[M], nxt[M], edge[M], tot, que[N], l, r, s, t, level[N], ans;
void init() {
tot = 1;
memset(head, 0, sizeof(head));
}
bool bfs() {
memset(level, 0, sizeof(level));
l = r = 1, que[l] = s, level[s] = 1;
while (l <= r) {
int cur = que[l++];
for (int i = head[cur]; i; i = nxt[i])
if (edge[i] && !level[ver[i]]) {
level[ver[i]] = level[cur] + 1, que[++r] = ver[i];
if (ver[i] == t)
return true;
}
}
return false;
}
int dinic(int cur, int increase) {
if (cur == t)
return increase;
int rest = increase;
for (int i = head[cur]; i; i = nxt[i])
if (edge[i] && level[ver[i]] == level[cur] + 1) {
int inc = dinic(ver[i], min(rest, edge[i]));
if (inc == 0) {
level[ver[i]] = 0;
continue;
}
rest -= inc, edge[i] -= inc, edge[i ^ 1] += inc;
}
return increase - rest;
}
int Dinic() {
int maxflow = 0, tmp;
while (bfs())
do {
tmp = dinic(s, 0x3f3f3f3f);
maxflow += tmp;
if (maxflow >= ans)
return n;
} while (tmp);
return maxflow;
}
void restore() {
for (int i = 1; i <= n + n; i++)
for (int j = head[i]; j; j = nxt[j])
if (!(j & 1))
edge[j] += edge[j ^ 1], edge[j ^ 1] = 0;
}
void addedge(int u, int v, int w) {
ver[++tot] = v, nxt[tot] = head[u], head[u] = tot, edge[tot] = w;
ver[++tot] = u, nxt[tot] = head[v], head[v] = tot, edge[tot] = 0;
}
int main() {
while (cin >> n >> m) {
init();
while (m--) {
int x, y;
scanf(" (%d,%d)", &x, &y);
x++, y++;
addedge(x, y + n, 0x3f3f3f3f), addedge(y, x + n, 0x3f3f3f3f);
}
for (int i = 1; i <= n; i++) addedge(i + n, i, 1);
ans = n;
for (s = 1; s <= n; s++)
for (t = n + 1; t < n + s; t++) {
restore();
ans = min(ans, Dinic());
}
cout << ans << endl;
}
return 0;
}
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