题目连接请戳

 

解题思路

比较容易想到的思路是开辟另一数组记录每个成员对应的敌人。

之后想明白每个团体只和一个团体敌对就行了。

 

代码

#include
     
      #include
      
       #define N 10010int fri[N], ene[N];int n;void make_set(int n){    for (int i = 1; i <= n; i++)        fri[i] = i;}int find(int x){    return (x == fri[x]) ? x : fri[x] = find(fri[x]);}int main(){    int c, a, b;    scanf("%d", &n);    make_set(n);    for (int i = 1; i <= n; i++) ene[i] = 0;    scanf("%d%d%d", &c, &a, &b);    while (!(c == 0 && a == 0 && b == 0)) {        a++; b++;        int fa = find(a);        int fb = find(b);        int qa = ene[fa], qb = ene[fb];        if (qa != 0) qa = find(qa);        if (qb != 0) qb = find(qb);        if (c == 1) {            if (qa == fb && qb == fa) printf("-1\n");            else {                if (qa != 0 && qb != 0) {                    fri[qa] = qb;                    fri[fa] = fb;                }                else if (qa != 0) fri[fb] = fa;                else if (qb != 0) fri[fa] = fb;                else fri[fa] = fb;            }        }        else if (c == 2) {            if (fa == fb) printf("-1\n");            else {                ene[fa] = fb;                ene[fb] = fa;                if (qa != 0) fri[qa] = fb;                if (qb != 0) fri[qb] = fa;            }        }        else if (c == 3) {            if (fa == fb) printf("1\n");            else printf("0\n");        }        else if (c == 4) {            if (qa == fb && qb == fa) printf("1\n");            else printf("0\n");        }        scanf("%d%d%d", &c, &a, &b);    }    return 0;}