巨型模拟题。
使用路径压缩合并并查集即可维护领土兼并功能。注意投放军队的时候也会产生冲突,但不输出任何信息。
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
using pii = pair<int, int>;
int n;
int m;
int k;
vector<string> names;
unordered_map<string, int> id;
vector<pii> pos;
vector<int> cntLand;
int q;
vector<int> p;
vector<vector<int>> grid;
int P(int a) {
if (p[a] == a) {
return a;
}
return p[a] = P(p[a]);
}
bool Cmp(int a, int b) {
return cntLand[a] < cntLand[b] || cntLand[a] == cntLand[b] &&
names[a] < names[b];
}
void Eat(int a, int b, bool show = true) {
p[b] = a;
id.erase(names[b]);
if (show) {
cntLand[a] += cntLand[b];
cout << names[a] << " wins!\n";
// cout << "eat " << names[b] << endl;
}
}
void Battle(int a, int b, bool show = true) {
if (Cmp(a, b)) {
Eat(b, a, show);
return;
}
Eat(a, b, show);
}
void Solve() {
cin >> n >> m >> k;
pos.resize(k);
p.resize(k);
names.resize(k);
grid.assign(n, vector<int>(m, -1));
cntLand.assign(k, 1);
for (int i = 0; i < k; i++) {
auto& str = names[i];
auto& [x, y] = pos[i];
cin >> str >> x >> y;
x--;
y--;
id[str] = i;
p[i] = i;
if (grid[x][y] != -1) {
Battle(P(grid[x][y]), i, false);
continue;
}
grid[x][y] = i;
}
cin >> q;
string str;
char c;
while (q--) {
cin >> str >> c;
auto it = id.find(str);
if (it == id.end()) {
cout << "unexisted empire.\n";
continue;
}
int idx = it->second;
auto [x, y] = pos[idx];
switch (c) {
case 'W':
x--;
break;
case 'S':
x++;
break;
case 'A':
y--;
break;
default:
y++;
break;
}
if (x < 0 || x >= n || y < 0 || y >= m) {
cout << "out of bounds!\n";
continue;
}
pos[idx] = {x, y};
int& t = grid[x][y];
// cout << x << ',' << y << ',' << t << endl;
if (t == -1) {
t = idx;
cntLand[idx]++;
cout << "vanquish!\n";
continue;
}
t = P(t);
// cout << t << ',' << idx << endl;
if (t == idx) {
cout << "peaceful.\n";
continue;
}
Battle(idx, t);
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
Solve();
return 0;
}

京公网安备 11010502036488号