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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
#62180#1869. Power Station of ArthinataRE 4ms60372kbC++145.8kb2022-11-17 16:43:422022-11-17 16:43:44

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-11-17 16:43:44]
  • 评测
  • 测评结果:RE
  • 用时:4ms
  • 内存:60372kb
  • [2022-11-17 16:43:42]
  • 提交

answer

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <ctime>
#include <cstdlib>
#include <set> 
#include <cassert>
#include <string>
#include <map>
#include <unordered_map>
#include <random>
#include <string>
#include <vector>
using namespace std;

#define ll long long
#define llu unsigned long long
#define mem(a, b) memset(a, b, sizeof(a))
#define el printf("\n")
#define pii pair<int, int>
#define af assert(0)
#define _(x) cout<<#x<<" = "<<x<<"\n"
#define __(x, y) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<"\n"
#define rep(i, l, r) for(int i = (l); i <= (r); ++i)
#define dep(i, l, r) for(int i = (l); i <= (r); ++i)

const int N = 1e6 + 10, M = 2e6 + 10;
const char Ans[20][20] = {"NO", "YES"};
int ver[M], nex[M], head[N], dcc_col[N], e_col[N], b[M * 2], col[2][N], val[2][N];
int book[2][3][2][M], sta[M], sta2[M];
char in[M];
vector<int> dcc[N];
int n, m, pos, cnt, ok, top, top2;

void init(){
    pos = cnt = ok = top = top2 = 0;
    for(int i = 0; i <= n; ++i){
        head[i] = dcc_col[i] = e_col[i] = 0;
        dcc[i].clear();
    }
}

void add(int x, int y){
    ver[++pos] = y, nex[pos] = head[x], head[x] = pos;
}

int find(int x){
    int l = 1, r = cnt;
    while(l < r){
        int mid = l + r >> 1;
        if(b[mid] >= x) r = mid;
        else l = mid + 1;
    }
    return l;
}

void dfs(int x, int c){
    dcc[c].push_back(x);
    dcc_col[x] = c;
    for(int i = head[x]; i; i = nex[i]){
        int y = ver[i];
        if(!dcc_col[y]){
            dfs(y, c);
        }
    }
}

void e_dfs(int x, int limit, int c){
    if(!ok) return;
    if(dcc_col[x] != limit) return;
    e_col[x] = c;
    for(int i = head[x]; i; i = nex[i]){
        int y = ver[i];
        if(e_col[y] == c){
            ok = 0;
            return;
        }
        else if(!e_col[y]){
            e_dfs(y, limit, 3 - c);
        }
    }
}

int check(int x){
    ok = 1;
    e_dfs(dcc[x][0], x, 1);
    if(!ok){ // 不是二分图
        top = top2 = 0;
        int A[2] = {0}, B[2] = {0};
        for(int i : dcc[x]){
            sta[++top] = val[0][i];
            ++A[col[0][i]], ++B[val[0][i]];
        }
        for(int i : dcc[x]){
            sta2[++top2] = val[1][i];
        }
        sort(sta + 1, sta + 1 + top), sort(sta2 + 1, sta2 + 1 + top);
        if(top != top2){
            return 0;
        }
        for(int i = 1; i <= top; ++i){
            if(sta[i] != sta2[i]){
                return 0;
            }
        }
        if(((1 & A[1]) ^ (1 & B[1])) == 0 && ((1 & A[0]) ^ (1 & B[0])) == 0){
            return 1;
        }
        return 0;
    }
    else{
        // printf("kfhd");
        top = 0;
        for(int k = 0; k < 2; ++k){
            for(int i : dcc[x]){
                sta[++top] = val[k][i];
                ++book[k][e_col[i]][col[k][i]][val[k][i]];
            }
        }
        // printf("sta : ");
        // for(int i = 1; i <= top; ++i){
        //     printf("%d ", sta[i]);
        // }
        // printf("\n");
        int flag = 1;
        for(int i = 1; i <= top; ++i){
            int now = sta[i];
            if(book[0][1][1][now] + book[0][2][0][now] != book[1][1][1][now] + book[1][2][0][now] ||
            book[0][1][0][now] + book[0][2][1][now] != book[1][1][0][now] + book[1][2][1][now]){
                flag = 0;
                break;
                // printf("129-1");
                // printf("%d %d %d %d\n", book[0][1][1][now], book[0][2][0][now], book[1][1][1][now], book[1][2][0][now]);
            }
            if(book[0][1][0][now] + book[0][1][1][now] + book[0][2][0][now] + book[0][2][1][now] != 
            book[1][1][0][now] + book[1][1][1][now] + book[1][2][0][now] + book[1][2][1][now]){
                flag = 0;
                break;
                // printf("133-1");
            }
        }

        for(int k = 0; k < 2; ++k){
            for(int i : dcc[x]){
                --book[k][e_col[i]][col[k][i]][val[k][i]];
            }
        }
        return flag;
    }
}

int main(){
    // freopen("1.in", "r", stdin);
    int t; scanf("%d", &t);
    for(int id = 1; id <= t; ++id){
        scanf("%d %d", &n, &m);
        init();
        for(int i = 1; i <= m; ++i){
            int x, y; scanf("%d %d", &x, &y);
            add(x, y), add(y, x);
        }
        for(int k = 0; k < 2; ++k){
            for(int i = 1; i <= n; ++i){
                scanf("%d", &val[k][i]);
                b[++cnt] = val[k][i];
            }
            scanf("%s", in + 1);
            for(int i = 1; i <= n; ++i){
                col[k][i] = (in[i] == 'B');
            }
        }

        sort(b + 1, b + 1 + cnt);
        int t = cnt; cnt = 0;
        for(int i = 1; i <= t; ++i){
            if(!cnt || b[i] != b[cnt]) b[++cnt] = b[i];
        }
        for(int k = 0; k < 2; ++k){
            for(int i = 1; i <= n; ++i){
                val[k][i] = find(val[k][i]);
            }
        }
        #ifdef DEBUG
            for(int i = 1; i <= n; ++i){
                printf("%d %d\n", val[0][i], col[0][i]);
            }
            el;
            for(int i = 1; i <= n; ++i){
                printf("%d %d\n", val[1][i], col[1][i]);
            }
            el;
        #endif
        cnt = 0;    
        for(int i = 1; i <= n; ++i){
            if(!dcc_col[i]){
                dfs(i, ++cnt);
            }
        }
        // for(int i = 1; i <= n; ++i){
        //     printf("%d ", dcc_col[i]);
        // }
        int ans = 1;
        for(int i = 1; i <= cnt; ++i){
            if(!check(i)){
                ans = 0;
                break;
            }
        }
        printf("%s\n", Ans[ans]);
        // break;
    }

    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 4ms
memory: 60372kb

input:

3
2 1
1 2
3 4
RR
4 3
BB
3 2
1 2
2 3
1 1 1
RBR
1 1 1
BBB
3 3
1 2
2 3
3 1
1 1 1
RBR
1 1 1
BBB

output:

YES
NO
YES

result:

ok 3 lines

Test #2:

score: -100
Dangerous Syscalls

input:

25054
5 10
4 5
4 2
4 3
4 1
5 2
5 3
5 1
2 3
2 1
3 1
12 2 9 10 7
RRRBB
10 2 9 7 12
RRBRB
1 0
4
R
4
R
8 4
4 3
1 5
8 5
6 5
7 2 11 10 9 6 3 5
RBRBBRBR
7 2 10 11 6 5 3 9
RBRBBRBR
7 4
7 2
5 1
5 6
5 4
12 5 9 14 5 12 12
RRBRBRR
14 12 9 5 12 12 5
RBBRBRB
10 1
4 6
5 3 2 9 7 3 11 11 6 8
RRRBRRBBBR
5 3 2 3 7 9 1...

output:


result: