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;
}
原文地址:https://www.cnblogs.com/gzygzy/p/10192531.html
时间: 2024-10-08 20:04:19