3.18-3.24
2.天梯赛练习集-L2部分
L2-001?紧急救援
比较裸的一个迪杰斯特拉，更新的时候需要考虑更多的信息
当由 x 更新结点 y 时。
d[y]更新时，到达被y的路径条数变为 x 的路径条数，y人数随之更新d[y] = d[x]+ dis[x][y]相等时，y结点路径条数增加，人数却要与之前的做比较，取一个最大值。
#include <bits/stdc++.h>
using namespace std;
const int N = 550,M = N*N;
const int inf = 0x3f3f3f3f;
int a[N],d[N],f[N],g[N],pa[N],n,m,s,e;
int vis[N];
int ver[M],edge[M],head[N],nxt[M],tot;
void add(int x,int y,int z){
ver[++tot] = y;
edge[tot] = z;
nxt[tot] = head[x];
head[x] = tot;
}
void print(int now){
if(now==pa[now])printf("%d",now-1);
else{
print(pa[now]);
printf(" %d",now-1);
}
}
int main(){
cin>>n>>m>>s>>e;
s++;e++;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
g[i] = a[i];
}
for(int i=1;i<=m;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
x++;y++;
if(x==y)continue;
add(x,y,z);
add(y,x,z);
}
for(int i=1;i<=n;i++)d[i] = inf;
d[s] = 0;f[s] = 1;g[s] = a[s],pa[s] = s;
for(int i=0;i<n;i++){
int id = inf;
for(int j=1;j<=n;j++)
if(!vis[j]&&(id==inf||d[j]<d[id]))id = j;
vis[id] = 1;
if(id==e)break;
for(int j=head[id];j;j = nxt[j]){
int y = ver[j];
if(vis[y])continue;
if(d[y] == d[id]+edge[j]){
f[y] += f[id];
if(g[y]<=g[id]+a[y]){
g[y] = g[id]+a[y];
pa[y] = id;
}
}
else if(d[y]>d[id]+edge[j]){
f[y] = f[id];
g[y] = a[y]+g[id];
d[y] = d[id]+edge[j];
pa[y] = id;
}
}
}
printf("%d %d\n",f[e],g[e]);
print(e);
puts("");
return 0;
}L2-007?家庭房产
- 一道挺恶心的并查集，匆匆过了
#include <bits/stdc++.h>
using namespace std;
const int N = 1010,M = 10010;
int fa[M],vis[M];
int find(int x){
return x==fa[x]?x:fa[x] = find(fa[x]);
}
vector<int> g[M];
struct node{
int num;
int id;
int cnt;
int flag;
double sum;
double ave;
double ave_num;
}a[M],b[M];
int n;
void merge(int x,int y){
x = find(x);
y = find(y);
if(x>y)swap(x,y);
fa[y] = x;
}
bool cmp(node a,node b){
if(a.ave==b.ave){
return a.id<b.id;
}
return a.ave>b.ave;
}
int main(){
cin>>n;
for(int i=0;i<=10000;i++){
fa[i] = i;
a[i].cnt = 1;
a[i].id = i;
}
for(int i=1;i<=n;i++){
int id,father,mather,k;
cin>>id>>father>>mather>>k;
vis[id] = 1;
if(father!=-1){
merge(id,father);
vis[father] = 1;
}
if(mather!=-1){
merge(id,mather);
vis[mather] = 1;
}
for(int j=0;j<k;j++){
int x;
cin>>x;
merge(id,x);
vis[x] = 1;
}
cin>>a[id].num>>a[id].sum;
}
for(int i=0;i<10000;i++){
if(vis[i]){
int x = find(i);
b[x].sum += a[i].sum;
b[x].num += a[i].num;
b[x].cnt ++;
b[x].flag = 1;
b[x].id = x;
}
}
int res = 0;
for(int i=0;i<10000;i++){
if(b[i].flag){
if(b[i].num){
b[i].ave = b[i].sum/b[i].cnt;
b[i].ave_num = b[i].num/b[i].cnt;
b[i].ave = (double)round(b[i].ave*1000.0)/1000;
}
else{
b[i].ave = 0;
b[i].ave_num = 0;
}
res++;
//printf("%d %d %d %lf\n",i,b[i].id,b[i].num,b[i].ave);
}
}
sort(b,b+10000,cmp);
cout<<res<<endl;
for(int i=0;i<res;i++){
printf("%04d %d %.3lf %.3lf\n",b[i].id,b[i].cnt,(double)b[i].num/b[i].cnt,b[i].ave);
}
return 0;
}L2-014
- 为了保证最后的出队序列是有序的，那么每个小队列也必须是有序的，最少的递减序列条数等于最长的单调不减序列的长度。所以直接求最长的单调不减序列长度即可。二分解法中每次覆盖之前的值，就可以理解为增加这一个位置纵向上的递减序列长度（比较抽象）
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N],dp[N];
int n;
int main(){
cin>>n;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
int len = 0;
for(int i=1;i<=n;i++){
int pos = lower_bound(dp,dp+len,a[i])-dp;
if(pos<len)dp[pos] = a[i];
else dp[len++] = a[i];
}
cout<<len<<endl;
return 0;
}L2-023?图着色问题
- 这个题很单纯，所用颜色必须为k个，多一个少一个都不行
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int a[N];
int n,m,k,t;
struct edge{
int x,y;
}eg[N*N];
bool check(){
for(int i=1;i<=m;i++){
if(a[eg[i].x]==a[eg[i].y])return false;
}
return true;
}
int main(){
cin>>n>>m>>k;
for(int i=1;i<=m;i++){
scanf("%d%d",&eg[i].x,&eg[i].y);
}
cin>>t;
set<int> st;
while(t--){
st.clear();
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
st.insert(a[i]);
}
//cout<<st.size()<<endl;
if(st.size()!=k){
puts("No");
}
else if(check()){
puts("Yes");
}
else puts("No");
}
return 0;
}3. 天梯赛练习集-L3部分
L3-001?凑零钱
- 记忆路径的背包问题，因为要满足答案字典序最小，所以先从大的选，也可以理解为尽可能让小的有更多的选择，使得答案中字典序小的优先选取
#include <bits/stdc++.h>
using namespace std;
int dp[10004][102];
int a[10004];
int main() {
ios::sync_with_stdio(false);
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1, a + n + 1, greater<int>()); // 保证字典序最小
dp[0][0] = 1;
for(int i = 1; i <= n; i++) {
for(int j = 0; j <= m; j++) {
dp[i][j] = dp[i - 1][j];
if(j >= a[i])
dp[i][j] += dp[i - 1][j - a[i]];
}
}
if(dp[n][m] == 0) {
cout << "No Solution";
} else {
vector<int> ans;
int i = n, j = m;
while(j > 0) {
if(j - a[i] >= 0 && dp[i - 1][j - a[i]]>0) {
ans.push_back(a[i]);
j -= a[i];
}
i--;
}
cout << ans[0];
for(int k = 1; k < ans.size(); k++) {
cout << " " << ans[k];
}
}
}L3-002?特殊堆栈
- 线段树
- 二分
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
//maintain the number of values between the l and r
struct SegmentTree{
int l,r;
int dat;
#define L(x) t[x].l
#define R(x) t[x].r
#define MID(x) (t[x].l+t[x].r)/2
}t[N*4];
int n;
void build(int p,int l,int r){
t[p].l = l;t[p].r = r;
t[p].dat = 0;
if(l==r)return;
build(p*2,l,MID(p));
build(p*2+1,MID(p)+1,r);
t[p].dat = t[p*2].dat+t[p*2+1].dat;
}
void change(int p,int x,int v){
if(L(p)==R(p)&&L(p)==x){
t[p].dat += v;
return ;
}
if(x<=MID(p)) change(p*2,x,v);
else change(p*2+1,x,v);
t[p].dat = t[p*2].dat+t[p*2+1].dat;
}
int query(int p,int x){
if(L(p)==R(p))return R(p);
if(x<=t[p*2].dat)return query(p*2,x);
else return query(p*2+1,x-t[p*2].dat);
}
int st[N];
char op[15];
int main(){
cin>>n;
int top = 0;
build(1,1,N-10);
while(n--){
scanf("%s",op);
if(op[1] == 'o'){
if(top==0)puts("Invalid");
else{
cout<<st[--top]<<endl;
change(1,st[top],-1);
}
}
else if(op[1]=='u'){
int x;
scanf("%d",&x);
st[top++] = x;
change(1,st[top-1],1);
//cout<<x<<endl;
}
else{
if(top==0)puts("Invalid");
else{
cout<<query(1,(top+1)/2)<<endl;
}
}
}
return 0;
}#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
vector<int> v;
int st[N],top;
int n,x;
char op[15];
int main(){
cin>>n;
while(n--){
scanf("%s",op);
if(op[1]=='o'){
if(top==0)puts("Invalid");
else{
printf("%d\n",st[--top]);
int p = lower_bound(v.begin(),v.end(),st[top])-v.begin();
v.erase(v.begin()+p);
}
}
else if(op[1]=='u'){
scanf("%d",&x);
st[top++] = x;
int p = lower_bound(v.begin(),v.end(),x)-v.begin();
v.insert(v.begin()+p,x);
}
else{
if(top==0)puts("Invalid");
else
printf("%d\n",v[(top-1)/2]);
}
}
return 0;
}L3-004?肿瘤诊断
- 三维BFS ，如果用DFS会爆栈
#include <bits/stdc++.h>
using namespace std;
int ma[1300][200][100];
int vis[1300][200][100];
int m,n,l,t;
int dx[6] = {1,-1,0,0,0,0};
int dy[6] = {0,0,1,-1,0,0};
int dz[6] = {0,0,0,0,1,-1};
bool check(int x,int y,int z){
if(x<1||x>m||y<1||y>n||z<1||z>l)return false;
if(vis[x][y][z]||ma[x][y][z]==0)return false;
return true;
}
struct node{
int x,y,z;
};
int bfs(int x,int y,int z){
queue<node> q;
int res = 0;
q.push({x,y,z});vis[x][y][z] = 1;
while(!q.empty()){
node u = q.front();q.pop();
res++;
for(int i=0;i<6;i++){
int a = u.x+dx[i];
int b = u.y+dy[i];
int c = u.z+dz[i];
if(check(a,b,c)){
q.push({a,b,c});
vis[a][b][c] = 1;
}
}
}
return res;
}
int main(){
//memset(vis,0,sizeof vis);
scanf("%d%d%d%d",&m,&n,&l,&t);
for(int i=1;i<=l;i++){
for(int x=1;x<=m;x++){
for(int y=1;y<=n;y++){
vis[x][y][i]=0;
scanf("%d",&ma[x][y][i]);
}
}
}
int res = 0;
for(int z=1;z<=l;z++){
for(int x=1;x<=m;x++){
for(int y=1;y<=n;y++){
if(ma[x][y][z]==1&&vis[x][y][z]==0){
int c = bfs(x,y,z);
//cout<<x<<endl;
if(c>=t)res+=c;
}
}
}
}
cout<<res<<endl;
return 0;
}L3-007?天梯地图
- 比较烦人的最短路，记录路径，比较路径，变量名上一定要细心
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
const int inf = 0x3f3f3f3f;
int len[N][N],tim[N][N];
int n,m;
int d[N],t[N],g[N],vis[N],st,ed;
vector<int> t_res,d_res;
void get_tim(int x){
if(x==st)t_res.push_back(x-1);
else {
get_tim(g[x]);
t_res.push_back(x-1);
}
}
void get_dis(int x){
if(x==st)d_res.push_back(x-1);
else{
get_dis(g[x]);
d_res.push_back(x-1);
}
}
bool check(){
if(d_res.size()!=t_res.size())return false;
for(int i=0;i<d_res.size();i++)
if(d_res[i] != t_res[i])return false;
return true;
}
int main(){
memset(len,0x3f,sizeof len);
memset(tim,0x3f,sizeof tim);
cin>>n>>m;
for(int i=1;i<=m;i++){
int x,y,one,l,t;
cin>>x>>y>>one>>l>>t;
x++;y++;
len[x][y] = l;
tim[x][y] = t;
if(one == 0){
len[y][x] = l;
tim[y][x] = t;
}
}
cin>>st>>ed;
st++;ed++;
for(int i=1;i<=n;i++)d[i] = inf,t[i] = inf,vis[i] = 0;
d[st] = 0;t[st] = 0;
priority_queue<pair<int,int> > q;
q.push(make_pair(0,st));
while(!q.empty()){
int x = q.top().second;q.pop();
if(vis[x])continue;
vis[x] = 1;
for(int i=1;i<=n;i++){
if(i==x)continue;
if(t[i] > t[x]+tim[x][i]){
t[i] = t[x]+tim[x][i];
d[i] = d[x]+len[x][i];
g[i] = x;
q.push(make_pair(-t[i],i));
}
else if(t[i]==t[x]+tim[x][i]&&d[i]>d[x]+len[x][i]){
d[i] = d[x]+len[x][i];
g[i] = x;
q.push(make_pair(-t[i],i));
}
}
}
printf("Time = %d",t[ed]);
get_tim(ed);
for(int i=1;i<=n;i++)d[i] = inf,t[i] = inf,vis[i] = 0;
d[st] = 0;t[st] = 1;
q.push(make_pair(0,st));
while(!q.empty()){
int x = q.top().second;q.pop();
if(vis[x])continue;
vis[x] = 1;
for(int i=1;i<=n;i++){
if(i==x)continue;
if(d[i] > d[x]+len[x][i]){
d[i] = d[x]+len[x][i];
t[i] = t[x]+1;
g[i] = x;
q.push(make_pair(-d[i],i));
}
else if(d[i] == d[x]+len[x][i]&&t[i] > t[x]+1){
t[i] = t[x]+1;
g[i] = x;
q.push(make_pair(-d[i],i));
}
}
}
get_dis(ed);
if(check()){
printf("; Distance = %d: %d",d[ed],d_res[0]);
for(int i=1;i<d_res.size();i++){
printf(" => %d",d_res[i]);
}
puts("");
}
else{
printf(": %d",t_res[0]);
for(int i=1;i<t_res.size();i++)printf(" => %d",t_res[i]);
puts("");
printf("Distance = %d: %d",d[ed],d_res[0]);
for(int i=1;i<d_res.size();i++){
printf(" => %d",d_res[i]);
}
puts("");
}
return 0;
}L3-016?二叉搜索树的结构
- BST 例题
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
struct node{
int l,r,data,fa;
}t[N];
map<int,int> id;
int a[N],n,m,d[N];
void insert(int &p,int fa,int now,int x,int dp){
if(p==-1){
p = now;
t[p].data = x;
t[p].fa = fa;
id[x] = now;
d[now] = dp;
return;
}
if(t[p].data>x)
insert(t[p].l,p,now,x,dp+1);
else insert(t[p].r,p,now,x,dp+1);
}
int main(){
scanf("%d",&n);
int root = -1;
for(int i=1;i<=n;i++)t[i].data=t[i].l=t[i].r= -1;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
insert(root,-1,i,a[i],1);
}
int q;cin>>q;
string st;int a,b;
while(q--){
cin>>a;
cin>>st;
if(st=="is"){
cin>>st>>st;
if(st=="root"){
if(!id.count(a)){
puts("No");continue;
}
if(id[a]==1)puts("Yes");
else puts("No");
}
else if(st=="parent"){
cin>>st>>b;
if(!id.count(a)||!id.count(b)){
puts("No");continue;
}
if(t[id[b]].fa == id[a])puts("Yes");
else puts("No");
}
else if(st=="left"){
cin>>st>>st>>b;
if(!id.count(a)||!id.count(b)){
puts("No");continue;
}
if(t[id[b]].l == id[a])puts("Yes");
else puts("No");
}
else if(st=="right"){
cin>>st>>st>>b;
if(!id.count(a)||!id.count(b)){
puts("No");continue;
}
if(t[id[b]].r==id[a])puts("Yes");
else puts("No");
}
}
else{
cin>>b;
cin>>st>>st;
if(st=="siblings"){
if(!id.count(a)||!id.count(b)||a==b){
puts("No");continue;
}
if(t[id[a]].fa==t[id[b]].fa)puts("Yes");
else puts("No");
}
else {
cin>>st>>st>>st;
if(!id.count(a)||!id.count(b)){
puts("No");continue;
}
if(d[id[a]] == d[id[b]])puts("Yes");
else puts("No");
}
}
}
return 0;
}L3-015?球队“食物链”
- 状压dp
#include <bits/stdc++.h>
using namespace std;
int n;
char s[22][22];
int d[1<<20][22];
int g[22];
bool check(int st){
int res = 0;
while(st&1)res++,st>>=1;
return res == n;
}
bool dp(int st,int fi,int now){
if(check(st)){
if(s[now][fi]=='W' || s[fi][now] == 'L'){
g[fi] = now;
return true;
}
else return false;
}
int flag = 0;
for(int i=0;i<n;i++){
if((st>>i)&1)continue;
if(s[i][fi] == 'W'||s[fi][i] == 'L'){
flag = 1;break;
}
}
if(flag == 0)return false;
for(int i=0;i<n;i++){
if((st>>i)&1)continue;
if(s[now][i]=='W' || s[i][now] == 'L'){
g[i] = now;
if(dp(st+(1<<i),fi,i))return true;
}
}
return false;
}
void print(int x,int fi){
if(x==fi)printf("%d",x+1);
else{
print(g[x],fi);
printf(" %d",x+1);
}
}
int main(){
scanf("%d",&n);
int num = 0;
for(int i=0;i<n;i++){
scanf("%s",s[i]);
}
for(int i=0;i<n;i++){
if(dp(1<<i,i,i)){
//for(int i=0;i<n;i++)printf("%d",g[i]);
print(g[i],i);
puts("");
return 0;
}
}
puts("No Solution");
return 0;
}L3-013?非常弹的球
- 很骚的一个题，有高中物理知识可以推导出，当角度为45度时，路程最远。
#include <cstdio>
int main() {
double w, p, v2, s = 0;
scanf("%lf%lf", &w, &p);
v2 = 2 * 1000 * 100 / w;
while (v2 > 0.000001) {
s += v2 / 9.8;
v2 *= (100 - p) * 0.01;
}
printf("%.3f", s);
return 0;
}原文地址：https://www.cnblogs.com/1625--H/p/10588573.html
时间： 2024-09-28 13:46:14