P5289 [十二省联考 2019] 皮配 做题记录

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题目 P5289 [十二省联考 2019] 皮配 - 洛谷 | 计算机科学教育新生态

参考题解 题解-十二省联考2019 皮配 - oier_hzy - 博客园

(第一次写这么难的题, 看着题解才勉强能写出来)

不考虑讨厌的导师, 超时的写法

设F[i][j] 为考虑完当前城市时, 选择蓝派系i人, 鸭派系j人的方案数

设两个辅助数组 P,G, P是如果当前城市选择蓝派系的方案数, G是当前城市选择红派系的方案数, citysum[zo]是当前城市的总人数,枚举当前城市中的每个学校, 如果当前学校是当前城市的第一个, 如果当前学校选择R派系P[i][j] += F[i - citysum[zo]][j],如果当前学校选择鸭派系P[i][j]+= F[i - citysum[zo]][j - s[it]] ,P[i][j] = F[i - citysum[zo]][j] + F[i - citysum[zo]][j - s[it]];  若当前学校不是城市中的第一个学校 P[i][j] = P[i][j] + P[i][j - s[it]]   使数组G = F, 对于数组G来说G[i][j] = G[i][j] + G[i][j - s[it]], 最后F[i][j] = P[i][j] + G[i][j]即为考虑完当前城市的方案数

最后整合答案, 因为选择红派系的人数不能超过c1, 即为i >= all - c1,i还要满足 i <= c0,同理,  all - d1 <= j <= d0

void solve(){
	int n, c;
	cin >> n >> c;
	int c0,c1,d0,d1;
	cin >> c0 >> c1 >> d0 >> d1;
	vector<int> s(n + 1), b(n + 1);
	vector<int> citysum(c + 1);
	vector<vector<int>> citysch(c + 1);
	int all = 0;
	for(int i = 1; i <= n; i++) cin >> b[i] >> s[i], citysum[b[i]] += s[i], citysch[b[i]].push_back(i), all += s[i];
	int k;
	cin >> k;
	
	while(k --){
		int x, y;
		cin >> x >> y;

	}
	vector<vector<int>> F(c0 + 1, vector<int> (d0 + 1));	
	F[0][0] = 1;
	for(int zo = 1; zo <= c; zo++) {
		if(citysch[zo].empty()) continue;

	
		vector<vector<int>> P(c0 + 1, vector<int> (d0 + 1)),G(c0 + 1, vector<int> (d0 + 1));
		for(auto it : citysch[zo]) {
			if(it == citysch[zo].front()) {
				for(int i = citysum[zo]; i <= c0; i++) {
					for(int j = 0; j <= d0; j++) {
						P[i][j] = F[i - citysum[zo]][j];
						if(j >= s[it]) P[i][j] = (P[i][j] + F[i - citysum[zo]][j - s[it]]) % mod;
					}
				}
				continue;
			}
			for(int i = 0; i <= c0; i++) {
				for(int j = d0; j >= s[it]; j--) P[i][j] = (P[i][j] + P[i][j - s[it]]) % mod;
				// 0;
			}
		}
		
		G = F;
		for(auto it : citysch[zo]) {
			
			for(int i = 0; i <= c0; i++) {
				for(int j = d0; j >= s[it]; j--) G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;			
			}
		}
		for(int i = 0; i <= c0; i++) {
			for(int j = 0; j <= d0; j++) F[i][j] = ((P[i][j] + G[i][j]) % mod + mod) % mod;
		}
	}

	int ans = 0;
	for(int i = max(0, all - c1); i <= c0; i++){
		for(int j = max(0,all - d1); j <= d0; j++) ans = (ans + F[i][j]) % mod/*, cout << F[i][j] << ' '*/;
	}
	cout << ans << endl;
}
考虑优化时间

一个学校选择是红蓝派系 , 鸭R派系是独立的, 所以分开考虑

设f[i]为选择蓝派系人数为i的方案数, g[j]为选择鸭派系为j的方案数

枚举每个城市 f[i] += f[i - citysum[zo]], 枚举每个学校 g[i] += g[i - s[j]]

对于f来说 all - c1<= i <= c0是满足条件的    g来说 all - d1 <= j <= d0 是满足条件的

答案即为(pre_f[c0] - pre_f[all - c1 - 1]) * (pre_g[d0] - pre_g[all - d0 - 1]);

考虑上讨厌的导师

记录每个城市是否有讨厌的导师, 每个学校是否有讨厌的导师

对于没有讨厌导师的城市 还用上一个优化时间的f的求法, 对于没有讨厌老师的学校, 也用g的求法

对于有讨厌的导师的城市, 采用第一次没有优化时间的写法

void solve1() {
	int n, c;
	cin >> n >> c;
	int c0,c1,d0,d1;
	cin >> c0 >> c1 >> d0 >> d1;
	vector<int> s(n + 1), b(n + 1), schhate(n + 1, -1);
	vector<int> citysum(c + 1), cityhate(n + 1);
	vector<vector<int>> citysch(c + 1);
	int all = 0;
	for(int i = 1; i <= n; i++) cin >> b[i] >> s[i], citysum[b[i]] += s[i], citysch[b[i]].push_back(i), all += s[i];
	int k;
	cin >> k;
	vector<int> f(c0 + 1), g(d0 + 1), pre_f(c0 + 1), pre_g(d0 + 1);
	pre_f[0] = f[0] = 1, pre_g[0] = g[0] = 1;
	while(k --){
		int x, y;
		cin >> x >> y;
		schhate[x] = y;
		cityhate[b[x]] = 1;
	}

	for(int i = 1; i <= n; i++) {
		if(schhate[i] == -1) {
			for(int j = d0; j >= s[i]; j--) g[j] = (g[j] + g[j - s[i]]) % mod;

		}
	}
	for(int i = 1; i <= d0; i++) pre_g[i] = (pre_g[i - 1] + g[i]) % mod;
	for(int i = 1; i <= c; i++) {
		if(!cityhate[i] && !citysch[i].empty()) {
			for(int j = c0; j >= citysum[i]; j--) f[j] = (f[j] + f[j - citysum[i]]) % mod;
		}
	}
	for(int i = 1; i <= c0; i++) pre_f[i] = (pre_f[i - 1] + f[i]) % mod;


	
	vector<vector<int>> F(c0 + 1, vector<int> (d0 + 1));	
	F[0][0] = 1;
	int Cs = 0, Ss = 0;
	for(int zo = 1; zo <= c; zo++) {
		if(citysch[zo].empty()) continue;
		if(!cityhate[zo]) continue;
		
		vector<vector<int>> P(c0 + 1, vector<int> (d0 + 1)),G(c0 + 1, vector<int> (d0 + 1));
		int cnt = 0;
		G = F;
		for(auto it : citysch[zo]) {
			if(schhate[it] == -1) continue;
			if(++cnt == 1) {
				if(!schhate[it]) {
					for(int i = c0; i >= citysum[zo]; i--) {
						for(int j = 0; j <= d0; j++) {
							P[i][j] =  F[i - citysum[zo]][j] % mod;
						}
					}
				}
				if(schhate[it] == 1)
				for(int i = c0; i >= citysum[zo]; i--) {
					for(int j = d0; j >= s[it]; j--) {
						P[i][j] =  F[i - citysum[zo]][j - s[it]] % mod;
					}
				}
			
				if(schhate[it] <= 1) {
					for(int i = 0; i <= c0; i++) {
						for(int j = d0; j >= s[it]; j--) {
							G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;
						
						}
					}
				}
				if(schhate[it] >= 2) {
					for(int i = c0; i >= citysum[zo]; i--) {
						for(int j = d0; j >= 0; j--) {
							P[i][j] = F[i - citysum[zo]][j];
							if(j >= s[it]) P[i][j] = (P[i][j] + F[i - citysum[zo]][j - s[it]]) % mod;
						}
					}
				}
			
				if(schhate[it] == 3) {
					for(int i = 0; i <= c0; i++) {
						for(int j = d0; j >= s[it]; j--) {
							G[i][j] = G[i][j - s[it]];
						}
						for(int j = 0; j < s[it]; j++) G[i][j] = 0;
 					}
				}
			
				continue;
			}


			if(schhate[it] == 1)
			for(int i = citysum[zo]; i <= c0; i++) {
				for(int j = d0; j >= s[it]; j--) {
					P[i][j] =  P[i][j - s[it]] % mod;
				}
				for(int j = 0; j < s[it]; j++) P[i][j] = 0;
			}
			if(schhate[it] <= 1) {
				for(int i = 0; i <= c0; i++) {
					for(int j = d0; j >= s[it]; j--) {
						G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] >= 2) {
				for(int i = citysum[zo]; i <= c0; i++) {
					for(int j = d0; j >= 0; j--) {
						if(j >= s[it]) P[i][j] = (P[i][j] + P[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] == 3) {
				for(int i = 0; i <= c0; i++) {
					for(int j = d0; j >= s[it]; j--) {
						G[i][j] = G[i][j - s[it]];
					}
					for(int j = 0; j < s[it]; j++) G[i][j] = 0;
				}
			}


		}
		
		
	
		for(int i = 0; i <= c0; i++) {
			for(int j = 0; j <= d0; j++) F[i][j] = ((P[i][j] + G[i][j]) % mod + mod) % mod;
		}
	}

	ll ans = 0;
	for(int i=0;i<=c0;++i)
	for(int j=0;j<=d0;++j) {
		int l1 = max(0, all - c1 - i), r1 = c0 - i; if(l1 > r1) continue;
		int l2 = max(0, all - d1 - j), r2 = d0 - j; if(l2 > r2) continue;
		int vf = pre_f[r1]; if(l1) vf += mod - pre_f[l1-1];
		int vg = pre_g[r2]; if(l2) vg += mod - pre_g[l2-1];
		

		ans = (ans + 1ll * vf * vg % mod * F[i][j] % mod) % mod;
	}

	cout << ans << endl;

}

有一个点TLE了, 卡一下常数, 每一次让G = F时间有点高, 每一次F数组并不是满的, 只需要记录到当前城市最多有几个人, 只要把不超过当前总人数的部分复制过去就行了

第一次AC代码

void solve1() {
	int n, c;
	cin >> n >> c;
	int c0,c1,d0,d1;
	cin >> c0 >> c1 >> d0 >> d1;
	vector<int> s(n + 1), b(n + 1), schhate(n + 1, -1);
	vector<int> citysum(c + 1), cityhate(n + 1);
	vector<vector<int>> citysch(c + 1);
	int all = 0;
	for(int i = 1; i <= n; i++) cin >> b[i] >> s[i], citysum[b[i]] += s[i], citysch[b[i]].push_back(i), all += s[i];
	int k;
	cin >> k;
	vector<int> f(c0 + 1), g(d0 + 1), pre_f(c0 + 1), pre_g(d0 + 1);
	pre_f[0] = f[0] = 1, pre_g[0] = g[0] = 1;
	while(k --){
		int x, y;
		cin >> x >> y;
		schhate[x] = y;
		cityhate[b[x]] = 1;
	}

	for(int i = 1; i <= n; i++) {
		if(schhate[i] == -1) {
			for(int j = d0; j >= s[i]; j--) g[j] = (g[j] + g[j - s[i]]) % mod;

		}
	}
	for(int i = 1; i <= d0; i++) pre_g[i] = (pre_g[i - 1] + g[i]) % mod;
	for(int i = 1; i <= c; i++) {
		if(!cityhate[i] && !citysch[i].empty()) {
			for(int j = c0; j >= citysum[i]; j--) f[j] = (f[j] + f[j - citysum[i]]) % mod;
		}
	}
	for(int i = 1; i <= c0; i++) pre_f[i] = (pre_f[i - 1] + f[i]) % mod;

	vector<vector<int>> F(c0 + 1, vector<int> (d0 + 1));	
	vector<vector<int>> P(c0 + 1, vector<int> (d0 + 1)),G(c0 + 1, vector<int> (d0 + 1));

	F[0][0] = 1;
	int Cs = 0, Ss = 0;
	for(int zo = 1; zo <= c; zo++) {
		if(citysch[zo].empty()) continue;
		if(!cityhate[zo]) continue;
		
		int cnt = 0;

		Cs += citysum[zo], Cs = min(Cs, c0);
		for(int i = 0; i <= Cs; i++) {
			for(int j = 0; j <= Ss; j++) G[i][j] = F[i][j];
		}
		for(int i = 0; i <= Cs; i++) {
			for(int j = 0; j <= Ss; j++) P[i][j] = 0;
		}
		for(auto it : citysch[zo]) {
			if(schhate[it] == -1) continue;
			int last = Ss;
			Ss += s[it], Ss = min(Ss, d0);
			if(++cnt == 1) {
				if(!schhate[it]) {
					for(int i = Cs; i >= citysum[zo]; i--) {
						for(int j = 0; j <= Ss; j++) {
							P[i][j] =  F[i - citysum[zo]][j] % mod;
						}
					}
				}

				if(schhate[it] == 1)
				for(int i = Cs; i >= citysum[zo]; i--) {
					for(int j = Ss; j >= s[it]; j--) {
						P[i][j] =  F[i - citysum[zo]][j - s[it]] % mod;
					}
				}
	
				if(schhate[it] <= 1) {
					for(int i = 0; i <= Cs; i++) {
						for(int j = Ss; j >= s[it]; j--) {
							G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;
							
						}
					}
				}
				if(schhate[it] >= 2) {
					for(int i = Cs; i >= citysum[zo]; i--) {
						for(int j = Ss; j >= 0; j--) {
							P[i][j] = F[i - citysum[zo]][j];
							if(j >= s[it]) P[i][j] = (P[i][j] + F[i - citysum[zo]][j - s[it]]) % mod;
						}
					}
				}

				if(schhate[it] == 3) {
					for(int i = 0; i <= Cs; i++) {
						for(int j = Ss; j >= s[it]; j--) {
							G[i][j] = G[i][j - s[it]];
						}
						for(int j = 0; j < s[it]; j++) G[i][j] = 0;
 					}
				}

				continue;
			}


			if(schhate[it] == 1)
			for(int i = citysum[zo]; i <= Cs; i++) {
				for(int j = Ss; j >= s[it]; j--) {
					P[i][j] =  P[i][j - s[it]] % mod;
				}
				for(int j = 0; j < s[it]; j++) P[i][j] = 0;
			}
			if(schhate[it] <= 1) {
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= s[it]; j--) {
						G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] >= 2) {
				for(int i = citysum[zo]; i <= Cs; i++) {
					for(int j = Ss; j >= 0; j--) {
						if(j >= s[it]) P[i][j] = (P[i][j] + P[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] == 3) {
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= s[it]; j--) {
						G[i][j] = G[i][j - s[it]];
					}
					for(int j = 0; j < s[it]; j++) G[i][j] = 0;
				}
			}


		}
		
		
		for(int i = 0; i <= Cs; i++) {
			for(int j = 0; j <= Ss; j++) F[i][j] = ((P[i][j] + G[i][j]) % mod + mod) % mod;
		}
	}

	ll ans = 0;
	for(int i=0;i<=Cs;++i)
		for(int j=0;j<=Ss;++j) {
			int l1 = max(0, all - i - c1), r1 = c0 - i;
			if(r1 < l1) continue;
			int l2 = max(0, all - j - d1), r2 = d0 - j;
			if(r2 < l2) continue;
			int a = pre_f[r1], b = pre_g[r2];
			if(l1) a = ((a - pre_f[l1 - 1]) % mod + mod) % mod;
			if(l2) b = ((b - pre_g[l2 - 1]) % mod + mod) % mod;
	
			ans += 1ll * a * b % mod * F[i][j] % mod;
			ans %= mod;
		}

	cout << ans << endl;

}

简化整合之后代码

void solve1() {
	int n, c;
	cin >> n >> c;
	int c0,c1,d0,d1;
	cin >> c0 >> c1 >> d0 >> d1;
	vector<int> s(n + 1), b(n + 1), schhate(n + 1, -1);
	vector<int> citysum(c + 1), cityhate(n + 1);
	vector<vector<int>> citysch(c + 1);
	int all = 0;
	for(int i = 1; i <= n; i++) cin >> b[i] >> s[i], citysum[b[i]] += s[i], citysch[b[i]].push_back(i), all += s[i];
	int k;
	cin >> k;
	vector<int> f(c0 + 1), g(d0 + 1), pre_f(c0 + 1), pre_g(d0 + 1);
	pre_f[0] = f[0] = 1, pre_g[0] = g[0] = 1;
	while(k --){
		int x, y;
		cin >> x >> y;
		schhate[x] = y;
		cityhate[b[x]] = 1;
	}

	for(int i = 1; i <= n; i++) {
		if(schhate[i] == -1) {
			for(int j = d0; j >= s[i]; j--) g[j] = (g[j] + g[j - s[i]]) % mod;

		}
	}
	for(int i = 1; i <= d0; i++) pre_g[i] = (pre_g[i - 1] + g[i]) % mod;
	for(int i = 1; i <= c; i++) {
		if(!cityhate[i] && !citysch[i].empty()) {
			for(int j = c0; j >= citysum[i]; j--) f[j] = (f[j] + f[j - citysum[i]]) % mod;
		}
	}
	for(int i = 1; i <= c0; i++) pre_f[i] = (pre_f[i - 1] + f[i]) % mod;

	vector<vector<int>> F(c0 + 1, vector<int> (d0 + 1));	
	vector<vector<int>> G(c0 + 1, vector<int> (d0 + 1));

	F[0][0] = 1;
	int Cs = 0, Ss = 0;
	for(int zo = 1; zo <= c; zo++) {
		if(citysch[zo].empty()) continue;
		if(!cityhate[zo]) continue;
		

		Cs += citysum[zo], Cs = min(Cs, c0);
		for(int i = 0; i <= Cs; i++) {
			for(int j = 0; j <= Ss; j++) G[i][j] = F[i][j];
		}
		
		for(auto it : citysch[zo]) {
			if(schhate[it] == -1) continue;
			Ss += s[it], Ss = min(Ss, d0);

			if(schhate[it] == 1)
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= s[it]; j--) {
						F[i][j] =  F[i][j - s[it]] % mod;
					}
					for(int j = 0; j < s[it]; j++) F[i][j] = 0;
				}
			if(schhate[it] <= 1) {
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= s[it]; j--) {
						G[i][j] = (G[i][j] + G[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] >= 2) {
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= 0; j--) {
						if(j >= s[it]) F[i][j] = (F[i][j] + F[i][j - s[it]]) % mod;
					}
				}
			}
			if(schhate[it] == 3) {
				for(int i = 0; i <= Cs; i++) {
					for(int j = Ss; j >= s[it]; j--) {
						G[i][j] = G[i][j - s[it]];
					}
					for(int j = 0; j < s[it]; j++) G[i][j] = 0;
				}
			}


		}
		for(int j = 0; j <= Ss; j++) {
			for(int i = Cs; i >= citysum[zo]; i--) F[i][j] = F[i - citysum[zo]][j];
			for(int i = 0; i < citysum[zo] && i <= Cs; i++) F[i][j] = 0;
		} 
		
		for(int i = 0; i <= Cs; i++) {
			for(int j = 0; j <= Ss; j++) F[i][j] = (F[i][j] + G[i][j]) % mod;
		}
	}

	ll ans = 0;
	for(int i=0;i<=Cs;++i)
		for(int j=0;j<=Ss;++j) {
			int l1 = max(0, all - i - c1), r1 = c0 - i;
			if(r1 < l1) continue;
			int l2 = max(0, all - j - d1), r2 = d0 - j;
			if(r2 < l2) continue;
			int a = pre_f[r1], b = pre_g[r2];
			if(l1) a = ((a - pre_f[l1 - 1]) % mod + mod) % mod;
			if(l2) b = ((b - pre_g[l2 - 1]) % mod + mod) % mod;
	
			ans += 1ll * a * b % mod * F[i][j] % mod;
			ans %= mod;
		}

	cout << ans << endl;

}

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