CodeForces AIM Tech Round 3 (Div. 1) 题解（CF708A，CF708B，CF708C，CF708D，CF708E）

A. Letters Cyclic Shift（CodeForces 708A）

题解：直接找到第一个不是a的位置i和i后面第一个为a的位置j(如果没有，j=len(s))，那么翻转的肯定就是[i,j]这一段。如果这个字符串都是a，那么就翻转最后一个字符。

#include <cstring>#include <cstdio>#include <algorithm>using namespace std;char s[100005];int main() {    scanf("%s", s + 1);    int n = strlen(s + 1);    bool flag = 0;    for (int i = 1; i <= n; i ++) {        if (!flag && s[i] != 'a') flag = 1;        if (flag && s[i] == 'a') {            for (int j = i; j <= n; j ++) printf("%c", s[j]);            break;        }        if (!flag && i != n) printf("%c", s[i]); else            printf("%c", (s[i] - 'a' - 1 + 26) % 26 + 'a');    }}

B. Recover the String（CodeForces 708B）

题解：由于知道了A[0][0]，假设0的数量是n0，那么A[0][0]肯定等于n0*(n0-1)/2，这样就可以算出n0，同理算出n1，同时判断A[0][0]和A[1][1]是否合法，同时也可以判断A[0][0]+A[0][1]+A[1][0]+A[1][1]是否等于(n0+n1)*(n0+n1-1)/2，即总数是否合法。最后根据A[0][1]的信息从前往后判断0是否可以填到第i位，满足0肯定肯定合法，同时1的限制肯定也合法了。还有一种特殊情况是当A[0][0]=A[0][1]=A[1][0]=0时n0等于0，1的情况同理。最后特判一下A[0][0]+A[0][1]+A[1][0]+A[1][1]=0的情况就可以了。

#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;typedef long long LL;LL a, b, c, d;char s[1000005];int main() {    scanf("%lld%lld%lld%lld", &a, &b, &c, &d);    if (a == 0 && b == 0 && c == 0 && d == 0) {printf("1\n"); return 0;}    LL num0 = sqrt(a * 2) + 1, num1 = sqrt(d * 2) + 1;    if (a == 0 && b == 0 && c == 0) num0 = 0;    if (b == 0 && c == 0 && d == 0) num1 = 0;    if (num0 * (num0 - 1) / 2 != a || num1 * (num1 - 1) / 2 != d) {printf("Impossible\n"); return 0;}    if (b + c != num0 * num1) {printf("Impossible\n"); return 0;}    LL all = num0 + num1;    LL e = all * (all - 1) / 2;    if (all > 1000000 || e != a + b + c + d) {printf("Impossible\n"); return 0;}    LL x = num0, y = num1, bb = b, cc = 0;    for (int i = 1; i <= all; i ++) {        if (y <= bb) s[i] = '0', bb -= y; else            s[i] = '1', y --;    }    printf("%s", s + 1);}

C. Centroids（CodeForces 708C）

题解：首先加入一个点能成为重心，那么就要存在节点数不大于n/2的子树。由于最多只有一个，我们只需要维护每个节点向下能切下来小于等于n/2的最大子树和向上能切下来小于等于n/2的最大子树。分别用一个树形dp来求出，最后判断每个节点最大的子树是否合法就可以了。

#include <cstring>#include <cstdio>#include <algorithm>#include <set>using namespace std;const int MAXN = 4e5 + 5;int n, size[MAXN], hev[MAXN], down[MAXN], up[MAXN];int tot, Last[MAXN], Go[MAXN * 2], Next[MAXN * 2];void link(int u, int v) {    Next[++ tot] = Last[u], Last[u] = tot, Go[tot] = v;}void dfs1(int now, int pre) {    size[now] = 1;    for (int p = Last[now]; p; p = Next[p]) {        int v = Go[p];        if (v == pre) continue;        dfs1(v, now);        size[now] += size[v];        down[now] = max(down[now], (size[v] <= n / 2) ? size[v] : down[v]);        hev[now] = max(hev[now], size[v]);    }}void dfs2(int now, int pre) {    multiset<int> d;    for (int p = Last[now]; p; p = Next[p]) {        int v = Go[p];        if (v != pre) d.insert((size[v] <= n / 2) ? size[v] : down[v]);    }    for (int p = Last[now]; p; p = Next[p]) {        int v = Go[p];        if (v == pre) continue;        if (n - size[v] <= n / 2) up[v] = max(up[v], n - size[v]); else {            up[v] = max(up[v], up[now]);            d.erase(d.find(size[v] <= n / 2 ? size[v] : down[v]));            if (!d.empty()) up[v] = max(up[v], *d.rbegin());            d.insert((size[v] <= n / 2) ? size[v] : down[v]);        }        dfs2(v, now);    }}int main() {    scanf("%d", &n);    for (int i = 1; i < n; i ++) {        int u, v;        scanf("%d%d", &u, &v);        link(u, v), link(v, u);    }    dfs1(1, 0);    dfs2(1, 0);    for (int i = 1; i <= n; i ++) {        int ans = 1;        if (n - size[i] > n / 2) ans = (n - size[i] - up[i] <= n / 2);         if (hev[i] > n / 2) ans = (hev[i] - down[i] <= n / 2);        printf("%d ", ans);    }}

D. Incorrect Flow（CodeForces 708D）

题解：设(u,v,f,w)表示从u到v有一条流量为f，费用为w的边。假设我们先连(n,1,Inf,0)，那么最终保证流量平衡就可以了。对于c>=f的边很好连，直接(u,v,c-f,1),(u,v,Inf,2),(v,u,f,1)，对于c

#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int Inf = 1e9;const int MAXN = 105, MAXM = 505;int n, m, S, T, ans, deg[MAXN], dis[MAXN], d[MAXN];int tot, Last[MAXN], Go[MAXM * 2], Next[MAXM * 2], Len[MAXM * 2], Cost[MAXM * 2];bool vis[MAXN], flag[MAXN];void open(int u, int v, int f, int w) {    Next[++ tot] = Last[u], Last[u] = tot, Go[tot] = v, Len[tot] = f, Cost[tot] = w;}void link(int u, int v, int f, int w) {    open(u, v, f, w), open(v, u, 0, -w);}int dfs(int now, int flow) {    if (now == T) {        ans += dis[T] * flow;        return flow;    }    int use = 0;    vis[now] = 1;    for (int p = Last[now]; p; p = Next[p]) {        int v = Go[p];        if (Len[p] && !vis[v] && dis[v] == dis[now] + Cost[p]) {            int t = dfs(v, min(flow - use, Len[p]));            Len[p] -= t, Len[p ^ 1] += t, use += t;            if (use == flow) return use;        }    }    return use;}int solve() {    while (1) {        memset(vis, 0, sizeof vis), memset(dis, 60, sizeof dis);        int l = 0, r = 1;        d[1] = S, dis[S] = 0, flag[S] = 1;        while (l != r) {            l = (l + 1) % MAXN;            int now = d[l];            for (int p = Last[now]; p; p = Next[p]) {                int v = Go[p];                if (Len[p] && dis[now] + Cost[p] < dis[v]) {                    dis[v] = dis[now] + Cost[p];                    if (!flag[v]) {                        flag[v] = 1;                        d[r = (r + 1) % MAXN] = v;                    }                }            }            flag[now] = 0;        }        if (dis[T] == dis[MAXN - 1]) return ans;        dfs(S, Inf);    }}int main() {    scanf("%d%d", &n, &m);    S = 0, T = n + 1, tot = 1;    link(n, 1, Inf, 0);    for (int i = 1; i <= m; i ++) {        int u, v, c, f;        scanf("%d%d%d%d", &u, &v, &c, &f);        deg[u] -= f, deg[v] += f;        if (c >= f) {            link(u, v, c - f, 1);            link(u, v, Inf, 2);            link(v, u, f, 1);        } else {            ans += f - c;            link(u, v, Inf, 2);            link(v, u, c, 1);            link(v, u, f - c, 0);        }    }    for (int i = 1; i <= n; i ++)        if (deg[i] < 0) link(i, T, -deg[i], 0); else link(S, i, deg[i], 0);    solve();    printf("%d\n", ans);}

E. Student’s Camp（CodeForces 708E）

施工中……