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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#578282#9319. Bull FarmmitthuRE 1ms8040kbC++143.2kb2024-09-20 17:57:542024-09-20 17:57:58

Judging History

你现在查看的是最新测评结果

  • [2024-09-20 17:57:58]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:8040kb
  • [2024-09-20 17:57:54]
  • 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[maxq];
struct node{
    int x, tim, num;
};
vector<pair<int, int> > edge[maxn];
vector<node> st[maxn];
priority_queue<pair<int, int>> q;
void dijkstra(int v0){
    for (int i = 1; i <= n; i++)
        dis[i] = inf, vis[i] = 0;
    dis[v0] = 0;
    q.push({0, v0});
    while (!q.empty()){
        int x = q.top().second;
        q.pop();
        if(vis[x])
            continue;
        vis[x] = 1;
        for(auto v : edge[x]){
            int y = v.first, w = v.second;
            //cout << x << " " << y << " " << w << endl;
            if(dis[y] > max(dis[x], w)){
                dis[y] = max(dis[x], w);
                if(!vis[y])
                    q.push({-dis[y], y});
            }
        }
    }
}
int find(int x){
    return (x == fa[x]) ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y, int num){
    x = find(x), y = find(y);
    if(x == y)
        return;
    edge[x].push_back({y, num});
    edge[y].push_back({x, num});
    fa[x] = y;
}
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();
    for (int i = 1; i <= Q; i++)
        st[i].clear();

    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');
        st[a].push_back({b, c, 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++)
                edge[j].push_back({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 (auto x : st[i]){
            if(dis[x.x] <= x.tim)
                ans[x.num] = 1;
            else
                ans[x.num] = 0;
        }
        //for (int j = 1; j <= n; j++)
            //cout << dis[j] << " ";
        //cout << endl;
    }
    for (int i = 1; i <= Q; i++)
        printf("%d", ans[i]);
    puts("");
}
int main(){

    scanf("%d", &T);
    for (; T;T--)
        solve();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 7956kb

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: 8040kb

input:

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

output:

010101

result:

ok single line: '010101'

Test #3:

score: -100
Runtime Error

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: