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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
#578323#9319. Bull FarmmitthuTL 127ms22184kbC++143.0kb2024-09-20 18:15:232024-09-20 18:15:24

Judging History

你现在查看的是最新测评结果

  • [2024-09-20 18:15:24]
  • 评测
  • 测评结果:TL
  • 用时:127ms
  • 内存:22184kb
  • [2024-09-20 18:15:23]
  • 提交

answer

#include<bits/stdc++.h>
const int maxn = 2e3 + 5;
const int maxm = 1e6 + 5;
const int maxq = 1e6 + 5;
const int inf = 1e9 + 7;
using namespace std;
int n, T, Q, l;
int t[maxn][maxn], in[maxn], fa[maxn];
char s[maxm];
int dis[maxn], vis[maxn], ans[maxn][maxn];
vector<pair<int, int> > edge[maxn];
queue<int> q[maxn];
void dijkstra(int v0){
    for (int i = 1; i <= n; i++)
        dis[i] = inf, vis[i] = 0;
    dis[v0] = 0;
    q[0].push(v0);
    for (int i = 0; i <= l; i++){
        while(!q[i].empty()){
            int x = q[i].front();
            q[i].pop();
            if(vis[x])
                continue;
            vis[x] = 1;
            for (auto v : edge[x]){
                int y = v.first, w = max(v.second, dis[x]);
                if(dis[y] > w){
                    dis[y] = w;
                    q[w].push(y);
                }
            }
        }
    }
}
int find(int x){
    return (x == fa[x]) ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y, int num){
    int fx = find(x), fy = find(y);
    if(x == y)
        return;
    edge[x].push_back({y, num});
    edge[y].push_back({x, num});
    fa[fx] = fy;
}
void solve(){
    scanf("%d %d %d", &n, &l, &Q);
    for (int i = 1; i <= l;i++){
        scanf("%s", s + 1);
        int m = strlen(s + 1);
        for (int j = 1, k = 1; j <= m; j += 2, k++)
            t[i][k] = (s[j] - '0') * 50 + (s[j + 1] - '0');
    }
    for (int i = 1; i <= n; i++)
        fa[i] = i, edge[i].clear(), fa[i] = i;
    for (int i = 1; i <= l;i++){
        for (int j = 1; j <= n; j++)
            in[j] = 0;
        for (int j = 1; j <= n; j++)
            in[t[i][j]]++;
        int sz = 0;
        for (int j = 1; j <= n;j++)
            sz += (in[j] > 0);
        //cout << i << "--- " << sz << endl;
        if(sz <= n-2)
            continue;
        if(sz == n)
            for (int j = 1; j <= n;j++)
                merge(j, t[i][j], i);
        else{
            int x = 0, y = 0;
            for (int j = 1; j <= n;j++){
                if(in[j] == 0)
                    x = j;
                else if(in[j] == 2)
                    y = j;
            }
            for (int j = 1; j <= n; j++){
                if(t[i][j] == y && j != x)
                    edge[j].push_back({x, i});
            }
        }
    }
    for (int i = 1; i <= n; i++){
        dijkstra(i);
        for (int j = 1; j <= n; j++)
            ans[i][j] = dis[j];
        //for (int j = 1; j <= n; j++)
            //cout << dis[j] << " ";
        //cout << endl;
    }
    for (int i = 1, a, b, c; i <= Q; i++)
    {
        scanf("%s", s + 1);
        a = (s[1] - '0') * 50 + (s[2] - '0');
        b = (s[3] - '0') * 50 + (s[4] - '0');
        c = (s[5] - '0') * 50 + (s[6] - '0');
        if(ans[a][b] <= c)
            printf("1");
        else
            printf("0");
    }
    puts("");
}
int main(){
    scanf("%d", &T);
    for (; T;T--)
        solve();
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 9172kb

input:

2
5 2 4
0305040201
0404040404
030300
020500
050102
020501
6 2 4
030603010601
010203060504
030202
060402
050602
060401

output:

1011
0100

result:

ok 2 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 7180kb

input:

1
3 3 6
020202
030301
030201
020102
030203
010201
010303
020303
010202

output:

010101

result:

ok single line: '010101'

Test #3:

score: 0
Accepted
time: 58ms
memory: 7484kb

input:

200
10 10 5000
01060:04020305080709
0103070:060204050908
09070503080401060:02
050308010204090:0607
03010502040607080:09
03080109020504060:07
06050:09040302080107
07080305010409060:02
030809010:0204060507
0:060908070201050304
060700
090:03
09080:
070405
010703
0:0100
080601
030600
070206
0:0:09
08040...

output:

011110001101101111111111111111111101111111110111011110110110111011010111111111111111111101111111111110111111110111111111111101111111111110111111111111111111110001100111111111111111111111111011101111111111111111111111111111111111111111011011110100111110111111110111111100111111101110111111111101111110...

result:

ok 200 lines

Test #4:

score: 0
Accepted
time: 127ms
memory: 22184kb

input:

1
2000 1 1000000
M=:]A@8UAY7W2JJ4KEHIA[HSCQ1ENSC`JXR;F3PJ:_@41P9Z=9HR8P<<:DUXRR9^WOQFL?NZP6S@=J0^WE32=6;\U0?88]Q_RNPUMT6YU<4<S]H?:7OCQYOT4YAV1^764ENWSDBED>M7A:BI>KSIR48JQ9B=N\5T3N4A2aF0@>3TI81<G7;YE>W`NMP<:IT4CI3D0=GZC3I\CLQJQBA9BDIS9SAM55KaVA<Z@D=>:Y?CQHUQ5U3a6UVI8OKX9_FAF^7=5M85;<0;8YPAM<7Z7PP7A=N...

output:

000101000101100010001000000010010110000001000001001100000000010000100001000000101100000000010000001000000001110000010110100000111100100000001000000000011001010001000001001000100000000100011001100110010000010000101100000011111000000010000101010010000000010101000001010111100000100000000000000101000100...

result:

ok single line: '000101000101100010001000000010...0101000101000000000010101001000'

Test #5:

score: -100
Time Limit Exceeded

input:

1
2000 2000 1000000
FVAaH7GRPO;_11Da5J18@3SMG==\G8E8S^6:M4L0JH>MN5IXT>2<7WJ3U8LVJS=;;3F13>0D0>VOIIU@EPHG6ABL6;K?T1PXDERLG07]5C9^GDKG<SBMIW;`4W8P3=469TIPKH0O34523_J5C2MJ17D25Z@=K8H@M>WK<CMK7EO]BPD7B6AW741J5YIHIa1:ERSG>L3N2^F3<4F`DLE@2AA5=8GZK6:192FB736[WMV7:^DA2C:<LK040VJBM3M]WXU50`407TR_?PLF@5VL7OSL...

output:


result: