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2021 RoboCom 世界机器人开发者大赛-本科组（复赛）
2022-07-06 04:43:00 【Zqchang】
官方题解
7-1 冒险者分队
分数 20
7-1 冒险者分队
一个莫名其妙的思维
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
// const int N =
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
int a, b, c, x, y, z;
signed main()
{
fast;
int t;
cin >> t;
while(t --)
{
cin >> a >> b >> c >> x >> y >> z;
if(a == x && b == y && c == z)
{
cout<<0<<endl;
continue;
}
int xx = x - a;
int yy = y - b;
int zz = z - c;
if((xx + yy + zz) || (xx % 20 != 0 || yy % 20 != 0 || zz % 20)) cout << -1 << endl;
else
{
int p[3] = {
xx, yy, zz};
sort(p, p + 3);
xx = p[0] / 20, yy = p[1] / 20, zz = p[2] / 20;
int ans = 0;
if((xx % 3 + 3) % 3 != (yy % 3 + 3) % 3 || (xx % 3 + 3) % 3 != (zz % 3 + 3) % 3)
{
cout<<-1<<endl;
continue;
}
// cout << xx << " " << yy << " " << zz << endl;
if(yy > 0)
{
ans += yy;
xx += 2 * yy;
zz -= yy;
yy = 0;
ans += zz / 3 * 2;
/* while(yy > 0) { xx += 2; yy -= 1; zz -= 1; ans ++; } while(xx < 0) { xx += 3; zz -= 3; ans += 2; } */
}
else if(yy < 0)
{
ans += abs(yy);
xx += abs(yy);
zz -= 2 * abs(yy);
yy = 0;
ans += zz / 3 * 2;
/* while(yy < 0) { xx += 1; yy += 1; zz -= 2; ans ++; } while(xx < 0) { xx += 3; zz -= 3; ans += 2; } */
}
else
{
ans += zz / 3 * 2;
// while(xx < 0)
// {
// xx += 3;
// zz -= 3;
// ans += 2;
// }
}
cout << ans << endl;
}
}
return 0;
}
7-2 拼题A打卡奖励
分数 25
7-2 拼题A打卡奖励
01背包的变形，在面临超时的情况下，明智的选择另一种作为限制
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a)
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int N = 1e3 + 10, M = 3e4 + 10;
int n, m;
int w[N], v[N];
int f[M];
signed main()
{
fast;
cin >> n >> m;
for(int i=1; i<=n; i++) cin >> v[i];//时间
for(int i=1; i<=n; i++) cin >> w[i];//金币
memset(f, 0x3f, sizeof f);
f[0] = 0;
for(int i=1; i<=n; i++)
for(int j=30000; j>=w[i]; j--)
{
f[j] = min(f[j], f[j - w[i]] + v[i]);
}
for(int i=30000; i>=0; i--)
{
if(f[i] <= m)
{
cout << i << endl;
break;
}
}
return 0;
}
7-3 快递装箱
分数 25
7-3 快递装箱
大模拟，没拿到满分，就十六分，不想改了，累了
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a)
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define LL long long
#define lowbit(p) p&(-p)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int N = 1e4 + 10;
// deque<int> a, b, c, d;
queue<PII> a, b, c;
deque<PII> d1, d;
int n, m, w1, w2, wm;
int f1[N], f2[N];
int res1, res2;
signed main()
{
// fast;
cin >> n >> wm >> w1 >> w2;
PII p;
for(int i=1; i<=n; i++)
{
while(d.size() >= 2)
{
p = d.back();
d.pop_back();
d1.push_back(p);
}
if(c.size())
{
p = c.front();
c.pop();
if(!d.size())
{
d.push_front(p);
f1[p.y] = 1;
}
else if(d.size())
{
PII q = d.front();
d.pop_front();
if(p.x + q.x <= wm && !f1[p.y]) d.push_front({
p.x + q.x, q.y});
else if((p.x + q.x > wm && !f1[p.y]) || f1[p.y])
{
if(q.x > w2) res2 ++;
else d.push_back(q);
d.push_front(p);
f1[p.y] = 1;
}
}
}
// else if(d.size() && i == n)
// {
// PII q = d.front();
// d.pop_front();
// if(q.x > w2) res2 ++;
// else d1.push_back(q);
// }
if(b.size())
{
p = b.front();
b.pop();
if(p.x > w2) res1 ++;
else c.push(p);
}
if(a.size())
{
p = a.front();
a.pop();
if(p.x > w1) f1[p.y] = 1;
b.push(p);
}
cin >> m;
if(d1.size())
{
p = d1.front();
d1.pop_front();
if(p.x + m <= wm)
{
a.push({
m + p.x, i});
f1[i] = 1;
}
else
{
d1.push_front(p);
a.push({
m, i});
}
}
else a.push({
m, i});
// cout << "di.size(): " << d1.size() << endl;
}
if(d.size())
{
p = d.front();
d.pop_front();
if(p.x > w2) res2 ++;
else d.push_front(p);
}
vector<int> v;
while(a.size())
{
p = a.front();
v.push_back(p.x);
a.pop();
// cout << "!!!" << endl;
}
while(b.size())
{
p = b.front();
v.push_back(p.x);
b.pop();
// cout << "!!!" << endl;
}
while(c.size())
{
p = c.front();
v.push_back(p.x);
c.pop();
// cout << "!!!" << endl;
}
while(d.size())
{
p = d.front();
v.push_back(p.x);
d.pop_front();
// cout << "!!!" << endl;
}
while(d1.size())
{
p = d1.front();
v.push_back(p.x);
d1.pop_front();
// cout << "!!!" << endl;
}
sort(v.begin(), v.end());
cout << res1 << " " << res2 << " " << v.size() << endl;
if(v.size() == 0) puts("None");
else
{
for(int i=0; i<v.size(); i++)
{
cout << v[i];
if(i == v.size() - 1) cout << endl;
else cout << " ";
}
}
return 0;
}
7-4 塔防游戏
分数 30
7-4 塔防游戏
头一次写二位最短路
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
#define PII pair<int, int>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define PIII pair<int, PII>
#define x first
#define y second
const int N = 110;
int dx[] = {
-1,1,0,0};
int dy[] = {
0,0,-1,1};//方向
struct node
{
int x, y, z;
};
int dist[N][N];
int f[110][110];
PII path[N][N];//记录僵尸路径
bool vis[N][N];
vector<node> v;//记录僵尸
int cnt[N][N];//记录走过的步数
vector<PII> p[4 * N];//记录每个僵尸的路线
int num[4 * N];//僵尸人数
map<PII, int> mp;
int now[N * 4];
int idx = 1;
int n, m, T;
int get(PII xx)
{
if(mp[xx] == 0) mp[xx] = idx++ ;//因为前面判断就是用的等于0，所以下标要从1开始
return mp[xx];
}
///*
void dij(PII u)
{
memset(dist, 0x3f, sizeof dist);
priority_queue<PIII, vector<PIII>, greater<PIII>> q;
dist[u.x][u.y] = 0;
q.push({
0, u});
while(q.size())
{
auto t = q.top();
q.pop();
if(vis[t.y.x][t.y.y]) continue;
vis[t.y.x][t.y.y] = 1;
auto tt = t.y;
for(int i=0; i<4; i++)
{
int xx = tt.x + dx[i];
int yy = tt.y + dy[i];
if(xx > n || xx < 1 || yy > m || yy < 1) continue;
if(dist[xx][yy] > dist[tt.x][tt.y] + f[xx][yy])
{
dist[xx][yy] = dist[tt.x][tt.y] + f[xx][yy];
cnt[xx][yy] = cnt[tt.x][tt.y] + 1;
path[xx][yy] = tt;
q.push({
dist[xx][yy], {
xx, yy}});
}
else if(dist[xx][yy] == dist[tt.x][tt.y] + f[xx][yy])
{
if(cnt[xx][yy] > cnt[tt.x][tt.y] + 1)
{
path[xx][yy] = tt;
q.push({
dist[xx][yy], {
xx, yy}});
}
}
}
}
}
//*/
signed main()
{
fast;
cin >> n >> m >> T;
PII start;
for(int i=0; i<=n+1; i++)
{
for(int j=0; j<=m+1; j++)
{
cin >> f[i][j];
if(i == 0 || j == 0 || i == n + 1 || j == m + 1)
{
if((i == 0 && j == 0) || (i == 0 && j == m+1) || (i == n+1 && j == 0) || (i == n+1 && j == m+1)) continue;
else v.push_back({
i, j, f[i][j]});
}
if(f[i][j] < 0) start = {
i, j};
}
}
// /*
dij(start);
// /*
for(auto i : v)
{
int id = get({
i.x, i.y});
num[id] = i.z;
PII kk;
for(int j = 0 ; j < 4 ; j++) //由僵尸边上走到地图里面去
if(i.x + dx[j] >= 1 && i.x + dx[j] <= n && i.y + dy[j] >= 1 && i.y + dy[j] <= m)
{
kk.x = i.x + dx[j] , kk.y = i.y + dy[j];
break ;
}
while(1)
{
p[id].push_back(kk);
if(kk.x == start.x && kk.y == start.y) break;//找到大本营的位置了
kk = path[kk.x][kk.y];
}
}
// /*
for(int t = 1; t <= T; t ++)
{
// memcpy()
set<PII> s;
for(auto i : v)
{
int id = get({
i.x, i.y});
if(num[id] == 0) continue;
int xx = p[id][now[id]].x;
int yy = p[id][now[id]].y;
if(f[xx][yy] > 0 && (xx != start.x || yy != start.y)) s.insert({
xx, yy});
else if(f[xx][yy] < 0 && (xx == start.x && yy == start.y)) s.insert({
xx, yy});
}
for(auto i : v)
{
int id = get({
i.x, i.y});
if(num[id] == 0) continue;
int xx = p[id][now[id]].x;
int yy = p[id][now[id]].y;
if(s.count({
xx, yy}) == 0) now[id] ++;//set中记录了哪些是炮台，所以不需要滚动数组优化
else
{
if(f[xx][yy] > 0) f[xx][yy] --;//减到0就没了，不减了，根据输出来搞得
else if(f[xx][yy] < 0) f[xx][yy]++;
num[id] --;
}
}
if(f[start.x][start.y] == 0) break;
}
// */
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
cout << f[i][j];
if(j < m) cout << " ";
else cout << endl;
}
if(f[start.x][start.y] == 0) cout << "Game Over" << endl;
return 0;
}
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