[JSOI2008]星球大战

Description

BZOJ1015

Solution

并差集经典应用。反向模拟，删点变加点，并差集维护联通块。

Code

 #include <cstdio>

const int N = 4e5+10;

int hd[N], nxt[N], to[N], cnt;
int fa[N];
int vis[N], a[N], ans[N];

inline void adde(int x, int y) {
    cnt++;
    nxt[cnt] = hd[x];
    to[cnt] = y;
    hd[x] = cnt;
}

int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

bool merge(int x, int y) {
    int fx = find(x), fy = find(y);
    if (fx == fy) return false;
    fa[fx] = fy;
    return true;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1, x, y; i <= m; ++i) {
        scanf("%d%d", &x, &y);
        x++, y++;
        adde(x, y);
        adde(y, x);
    }
    for (int i = 1; i <= n; ++i) fa[i] = i;
    int k, cnt = n;
    scanf("%d", &k);
    for (int i = 1; i <= k; ++i) scanf("%d", &a[i]), vis[++a[i]] = 1, cnt--;
    for (int i = 1; i <= n; ++i) if (!vis[i]) {
        for (int j = hd[i]; j; j = nxt[j]) if (!vis[to[j]]) {
            if (merge(i, to[j])) cnt--;
        }
    }
    ans[k+1] = cnt;
    for (int i = k; i; --i) {
        cnt++;
        vis[a[i]] = 0;  
        for (int j = hd[a[i]]; j; j = nxt[j]) if (!vis[to[j]]) {
            if (merge(a[i], to[j])) cnt--;
        }
        ans[i] = cnt;
    }
    for (int i = 1; i <= k+1; ++i) {
        printf("%d\n", ans[i]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/wyxwyx/p/bzoj1015.html