CF623A Graph and String

Posted 2021-01-30 gzygzy

CF623A Graph and String

休闲的题目.
WA了5次,hack数据爽的一批.
考虑补图.
显然,补图中，出度为0的点是B.
然后二分图染色.
如果不是二分图,就输出"No"
染完色后，看看是否满足数据即可
写的太丑了,捂脸.

/*header*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#define rep(i , x, p) for(int i = x;i <= p;++ i)
#define sep(i , x, p) for(int i = x;i >= p;-- i)
#define gc getchar()
#define pc putchar
#define ll long long
#define mk make_pair
#define fi first
#define se second
using std::min;
using std::max;
using std::swap;
const int maxN = 500 + 7;

inline int gi() {
    int x = 0,f = 1;char c = gc;
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}return x * f;
}

void print(int x) {
    if(x < 0) pc('-') , x = -x;
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
}

int n , m , dis[maxN][maxN], col[maxN];
bool flag;

void dfs(int now , int cl) {
    col[now] = cl;
    rep(i , 1, n) {
        int v = i;
        if(!dis[now][v]) continue;
        if(col[v] == col[now]) {
            puts("No");
            exit(0);
        }
        if(!col[v]) dfs(v , 3 - cl);
    }
}

int main() {
    n = gi();m = gi();
    rep(i , 1, n) rep(j , 1, n) dis[i][j] = 1;
    rep(i , 1, n) dis[i][i] = 0;
    while(m --) {
        int u = gi() , v = gi();
        dis[v][u] = dis[u][v] = 0;
    }
    int B = 0;bool fl = false;
    for(int i = 1;i <= n;++ i) {
        B = i;
        for(int j = 1;j <= n;++ j) 
            if(dis[i][j]) fl = true;
        if(fl) break;
    }
    if(fl) dfs(B , 1);
    rep(i , 1, n) {
        rep(j , 1, i - 1) {
            if(!dis[i][j]) {
                if(col[i] + col[j] == 3) {
                    puts("No");return 0;
                    }
                }
            }
        }
        rep(i , 1, n) {
            if(col[i] == 0) {
                rep(j , 1, n) {
                    if(dis[i][j]) {
                        puts("No");return 0;
                    }
                }
            }
        }
    puts("Yes");
    for(int i = 1;i <= n;++ i) {
        if(col[i] == 0) pc('b');
        if(col[i] == 1) pc('a');
        if(col[i] == 2) pc('c');
    }
    return 0;
}