P1196 [NOI2002]银河英雄传说 【带权并查集】

 

 

 思路

　　用sum记录每个舰队的战舰数量， tohead 记录当前舰离舰首的距离，那么求任意两舰之间有多少舰显然就是 abs( tohead[i] - tohead[j] ) - 1；

CODE

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int maxn = 5e4 + 7;

int fa[maxn];
int tohead[maxn], sum[maxn];
int t;

int fid(int x) {
    if(fa[x] == x) {
        return x;
    }
    int r1 = fa[x], r2 = fid(fa[x]);
    fa[x] = r2;
    tohead[x] += tohead[r1];
    return r2;
}

void join(int x, int y) {
    int fx = fid(x), fy = fid(y);
    fa[fy] = fx;
    tohead[fy] = sum[fx];
    sum[fx] += sum[fy];
    sum[fy] = 0;
}

int main()
{
    scanf("%d",&t);
    for ( int i = 1; i <= 30007; ++i ) {
        fa[i] = i;
        sum[i] = 1;
    }
    while(t--) {
        char s[5];
        int x, y;
        cin >> s;
        scanf("%d %d",&x, &y);
        if(s[0] == 'M') {
            join(x, y);
        }
        else {
            if(fid(x) != fid(y)) {
                printf("-1
");
                continue;
            }
            else {
                printf("%d
",abs(tohead[x] - tohead[y]) - 1);
            }
        }
    }
    return 0;
}