题目链接
A. Vasya and Chocolate
题意
已知钱，价格，赠送规则求最多获得巧克力数
思路
常规算即可
代码
#include
#define DBG(x) cerr << #x << " = " << x << endl;
using namespace std;
typedef long long LL;
LL t,s,a,b,c;
int main(){
cin >> t;
while(t--){
cin >> s >> a >> b >> c;
LL ans=s/c;
ans+=(ans/a)*b;
cout << ans << endl;
}
return 0;
}
B. Vasya and Isolated Vertices
题意
给出无向图点和边数问最多和最少孤立点的数量
思路
使孤立点尽可能少就让一条边尽可能消去两个点,否则让其尽可能消去一个点
代码
#include
#define DBG(x) cerr << #x << " = " << x << endl;
using namespace std;
typedef long long LL;
LL n,m,i;
int main(){
scanf("%I64d%I64d",&n,&m);
while(i*(i-1)/2 < m)i++;
printf("%I64d %I64d\n",(m*2 < n) ? n-m*2 : 0,n-i);
return 0;
}
C. Make It Equal
题意
给出一些块柱,每次移动一层及以上所有块,在一次不移动超过k的前提下使所有柱高度一致的最少次数
思路
枚举当前移动的高度,使被移动的块尽可能接近k,需要预处理每层会影响的块数,枚举高度二分找位置,维护后缀即可完成预处理
代码
#include
#define DBG(x) cerr << #x << " = " << x << endl;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5+5;
using namespace std;
int n,k,a[maxn],maxH=-inf;
int b[maxn];
vectorvec;
int main(){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
sort(a+1,a+1+n);
for(int i=2;i<=n;i++){
int det=a[i]-a[1];
maxH=max(maxH,det);
if(det)vec.push_back(det);
}
int siz=vec.size();
for(int i=maxH;i>=1;i--){
b[i]=lower_bound(vec.begin(),vec.end(),i)-vec.begin()+1;
b[i]=siz-b[i]+1;
b[i]+=b[i+1];
}
int now=0,cnt=0;
for(int i=maxH;i>=1;i--){
if(b[i]-now > k)now=b[i+1],cnt++;
}
if(now != b[1] && b[1]-now <= k)cnt++;
printf("%d\n",cnt);
return 0;
}
D. Three Pieces
题意
给定棋盘和三颗棋子,问遍历棋盘的最少步数即满足最少步数前提下替换棋子的最少次数
思路
从起点开始大力搜索,枚举所有可能的后继状态
代码
#include
#define DBG(x) cerr << #x << " = " << x << endl;
const int maxn = 15;
const int maxm = 205;
const int inf = 0x3f3f3f3f;
using namespace std;
int n,k;
int mp[maxn][maxn],vis[maxn][maxn],dp[maxn][maxn][3][maxm][maxm];
int sx,sy,ex,ey;
int dx1[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};
int dx2[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
int dx3[4][2]={{-1,-1},{-1,1},{1,-1},{1,1}};
int ans=inf;
struct node{
int r,c,who,time,pre;
node(int _r,int _c,int _who,int _time,int _pre){
r=_r,c=_c,who=_who,time=_time,pre=_pre;
}
};
void bfs(int sx, int sy) {
memset(dp, -1, sizeof dp);
dp[sx][sy][0][0][1] = dp[sx][sy][1][0][1] = dp[sx][sy][2][0][1] = 0;
queueq;
q.push(node(sx, sy, 0, 0, 1));
q.push(node(sx, sy, 1, 0, 1));
q.push(node(sx, sy, 2, 0, 1));
while(!q.empty()) {
node nd = q.front();
q.pop();
int x = nd.r, y = nd.c, z = nd.who, t = nd.time, k = nd.pre;
for(int i = 0; i < 3; i++) {
if(i == z) continue;
if(dp[x][y][i][t + 1][k] != -1) continue;
dp[x][y][i][t + 1][k] = dp[x][y][z][t][k] + 1;
q.push(node(x, y, i, t + 1, k));
}
if(z == 0) {
for(int i = 0; i < 8; i++) {
int nx=x+dx1[i][0];
int ny=y+dx1[i][1];
int nk=k;
if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
if(mp[nx][ny] == k + 1) nk++;
if(dp[nx][ny][z][t][nk] != -1) continue;
dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
q.push(node(nx, ny, z, t, nk));
}
}
if(z == 1) {
for(int i = 0; i < 4; i++) {
for(int j = 1;j <= 10;j++) {
int nx=x+j*dx2[i][0];
int ny=y+j*dx2[i][1];
int nk=k;
if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
if(mp[nx][ny] == k + 1) nk++;
if(dp[nx][ny][z][t][nk] != -1) continue;
dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
q.push(node(nx, ny, z, t, nk));
}
}
}
if(z == 2) {
for(int i = 0; i < 4; i++) {
for(int j = 1;j <= 10;j++) {
int nx=x+j*dx3[i][0];
int ny=y+j*dx3[i][1];
int nk=k;
if(nx < 1 || nx > n || ny < 1 || ny > n) continue;
if(mp[nx][ny] == k + 1) nk++;
if(dp[nx][ny][z][t][nk] != -1) continue;
dp[nx][ny][z][t][nk] = dp[x][y][z][t][k] + 1;
q.push(node(nx, ny, z, t, nk));
}
}
}
}
}
int main(){///0 == knight,1 == bishop,2 == rook
scanf("%d",&n);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
scanf("%d",&mp[i][j]);
if(mp[i][j] == 1){sx=i;sy=j;}
if(mp[i][j] == n*n){ex=i;ey=j;}
}
}
bfs(sx,sy);
for(int i=0;i E. Side Transmutations
看这篇
F. Up and Down the Tree
看这篇