QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#508823#7671. Metal Processing PlantKiharaToumaTL 1ms5936kbC++232.7kb2024-08-07 20:29:362024-08-07 20:29:37

Judging History

你现在查看的是最新测评结果

  • [2024-08-07 20:29:37]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:5936kb
  • [2024-08-07 20:29:36]
  • 提交

answer

//qoj7671
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 410;
int n, e[N][N], tot, ans = 2e9 + 10, fa[N], siz[N], col[N];
vector<int> son[N];
struct edge{
	int val, x, y;
} eg[N*N];

bool cmp(edge x, edge y){
	return x.val > y.val;
}
int gf(int x){
	return fa[x] == x ? x : gf(fa[x]);
}
void dfs(int x, int cl){
	col[x] = cl;
	for(int i : son[x]){
		dfs(i, 1-cl);
	}
}
void mg(int x, int y){
	if(siz[x] > siz[y]) swap(x, y);
	son[y].push_back(x);
	siz[y] += siz[x];
	dfs(x, 1 - col[y]);
}

int dfn[N], low[N], scc[N], sct, cnt, ins[N], st[N], tp;
vector<int> g[N];
void add(int x, int y){
	g[x].push_back(y);
}
void tg(int x){
	dfn[x] = low[x] = ++ cnt;
	ins[x] = 1;
	st[++tp] = x;
	for(int i : g[x]){
		if(!dfn[i]){
			tg(i);
			low[x] = min(low[x], low[i]);
		} else if(ins[i]){
			low[x] = min(low[x], dfn[i]);
		}
	}
	if(dfn[x] == low[x]){
		++ sct;
		while(st[tp] != x){
			scc[st[tp]] = sct;
			ins[st[tp]] = 0;
			-- tp;
		}
		scc[st[tp]] = sct;
		ins[st[tp]] = 0;
		-- tp;
	}
}

bool chk(int p, int q){
	sct = cnt = tp = 0;
	for(int i = 1; i <= n + n; ++ i){
		vector<int> ().swap(g[i]);
		dfn[i] = low[i] = scc[i] = ins[i] = st[i] = 0;
	}
	for(int i = 1; i <= tot; ++ i){
		int x = eg[i].x, y = eg[i].y;
		if(i == p){
			add(x + n, x);
			add(y + n, y);
		} else if(i < p){
			add(x, y + n);
			add(x + n, y);
			add(y, x + n);
			add(y + n, x);
		} else if(eg[i].val > q){
			add(x + n, y);
			add(y + n, x);
		}
	}
	for(int i = 1; i <= n + n; ++ i){
		if(!dfn[i]){
			tg(i);
		}
	}
	for(int i = 1; i <= n; ++ i){
		if(scc[i] == scc[i+n]){
			return 0;
		}
	}
	return 1;
}

int main(){
	scanf("%d", &n);
	int mx = 0;
	for(int i = 1; i <= n; ++ i){
		fa[i] = i;
		siz[i] = 1;
		col[i] = 0;
		for(int j = i + 1; j <= n; ++ j){
			scanf("%d", &e[i][j]);
			mx = max(mx, e[i][j]);
			e[j][i] = e[i][j];
			eg[++tot] = (edge){ e[i][j], i, j };
		}
	}
	sort(eg + 1, eg + tot + 1, cmp);
	for(int i = 1; i <= tot; ++ i){
		int flg = 0;
		if(gf(eg[i].x) == gf(eg[i].y)){
			if(col[eg[i].x] == col[eg[i].y]){
				flg = 1;
			} else {
				continue;
			}
		}
		mg(eg[i].x, eg[i].y);
		int l = i + 1, r = tot;
		while(l < r){
			int mid = l + r + 1 >> 1;
			if(chk(i, eg[mid].val)){
				l = mid;
			} else {
				r = mid - 1;
			}
		}
		if(chk(i, l)){
			ans = min(ans, eg[i].val + eg[l].val);
		}
		if(flg){
			break;
		}
	}
	ans = min(ans, mx);
	for(int i = 1; i <= n; ++ i){
		mx = 0;
		for(int j = 1; j <= n; ++ j){
			for(int k = 1; k <= n; ++ k){
				if(j != i && k != i){
					mx = max(mx, e[j][k]);
				}
			}
		}
		ans = min(ans, mx);
	}
	printf("%d\n", ans);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 5936kb

input:

5
4 5 0 2
1 3 7
2 0
4

output:

4

result:

ok single line: '4'

Test #2:

score: 0
Accepted
time: 1ms
memory: 5828kb

input:

7
1 10 5 5 5 5
5 10 5 5 5
100 100 5 5
10 5 5
98 99
3

output:

15

result:

ok single line: '15'

Test #3:

score: 0
Accepted
time: 0ms
memory: 3880kb

input:

1

output:

0

result:

ok single line: '0'

Test #4:

score: 0
Accepted
time: 0ms
memory: 3792kb

input:

2
1

output:

0

result:

ok single line: '0'

Test #5:

score: 0
Accepted
time: 1ms
memory: 5884kb

input:

2
1000000000

output:

0

result:

ok single line: '0'

Test #6:

score: 0
Accepted
time: 0ms
memory: 3788kb

input:

3
1000000000 1000000000
1000000000

output:

1000000000

result:

ok single line: '1000000000'

Test #7:

score: 0
Accepted
time: 0ms
memory: 3836kb

input:

4
1000000000 1000000000 1000000000
1000000000 1000000000
1000000000

output:

1000000000

result:

ok single line: '1000000000'

Test #8:

score: 0
Accepted
time: 0ms
memory: 3980kb

input:

3
78 97
24

output:

24

result:

ok single line: '24'

Test #9:

score: -100
Time Limit Exceeded

input:

200
202018752 202018795 202018793 100018905 202018758 202018741 202018727 202018766 202018728 100018879 202018781 100018860 202018785 100018841 100018910 100018836 100018883 100018847 202018734 202018775 100018830 100018901 202018773 202018754 202018737 100018843 202018788 202018778 202018777 202018...

output:


result: