Codeforces Round #294 (Div. 2)

D. A and B and Interesting Substrings

题意：给定26个小写字母的权值和一个字符串， 求出该字符串有多少个连续的子串，首尾相同， 并且除了首尾权值和为0。

思路：权值和为0说明子串的两个端点的前缀和相同， 先预处理出前缀和， 然后因为首尾要相同， 所以开26个map记录一个某个字符的前缀和的出现次数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>using namespace std;#define LL long long#define inf 0x3f3f3f3f#define eps 1e-8#define ULL unsigned long long#define mxn 100020#define mxe 200020#define mxnode 20020#define mod 1000000007#define vi vector<int>#define pii pair<int, int>#define vii vector<pii >LL sum[mxn];int a[30];char s[mxn];map<LL, int> b[30];int main() {    while(1) {        for(int i = 0; i < 26; ++i) scanf("%d", &a[i]);        scanf("%s", s);        int n = strlen(s);        for(int i = 0; s[i]; ++i)            sum[i] = i == 0? a[s[i]-'a']: sum[i-1] + a[s[i]-'a'];        LL ans = 0;        for(int i = 0; i < 30; ++i) b[i].clear();        for(int i = 0; s[i]; ++i) {            int c = s[i] - 'a';            ans += b[c][sum[i-1]];            ++b[c][sum[i]];        }        printf("%I64d\n", ans);        return 0;    }    return 0;}

E.  A and B and Lecture Rooms

题意：给定一棵树，和m个询问， 每个询问包括两个节点u和v，要求求出树上跟到u和v距离相同的节点的个数。

思路： 先求出u到v的距离dis， 可以用lca求， 如果dis==0的话， 答案就是n， 如果dis为奇数的话， 答案就是0.

最后，如果dis是偶数， 假设u的深度小于v的深度， 则求出v的dis/2级祖先w， 如果w和p相同的话， v的dis/2-1级祖先是x， u的dis/2-1级祖先是y， 答案就是n-sz[x]-sz[y],

如果w和p不相同的话， 答案就是sz[w] - sz[x].

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>using namespace std;#define LL long long#define inf 0x3f3f3f3f#define eps 1e-8#define ULL unsigned long long#define mxn 100020#define mxe 200020#define mxnode 20020#define mod 1000000007#define vi vector<int>#define pii pair<int, int>#define vii vector<pii >int fst[mxn], nxt[mxe], to[mxe], e;void init() {    memset(fst, -1, sizeof fst);    e = 0;}void add(int u, int v) {    to[e] = v, nxt[e] = fst[u], fst[u] = e++;}int n, m;int sz[mxn];int fa[22][mxn], dep[mxn];void dfs(int u, int p, int d) {    sz[u] = 1;    dep[u] = d;    fa[0][u] = p;    for(int i = fst[u]; ~i; i = nxt[i]) {        int v = to[i];        if(v == p) continue;        dfs(v, u, d + 1);        sz[u] += sz[v];    }}void lcaInit() {    for(int k = 0; k < 21; ++k) {        for(int i = 1; i <= n; ++i) {            if(fa[k][i] != -1)                fa[k+1][i] = fa[k][fa[k][i]];            else                fa[k+1][i] = -1;        }    }}int lca(int u, int v) {    if(dep[u] > dep[v]) swap(u, v);    for(int k = 0; k < 22; ++k) {        if((dep[v] - dep[u]) >> k & 1)            v = fa[k][v];    }    if(u == v) return u;    for(int k = 21; k >= 0; --k) {        if(fa[k][u] != fa[k][v])            u = fa[k][u], v = fa[k][v];    }    return fa[0][u];}int kfa(int u, int k) {    int t = u;    for(int i = 0; i < 22; ++i) {        if(k >> i & 1)            u = fa[i][u];    }    return u;}int main() {//  freopen("tt.txt", "r", stdin);    while(scanf("%d", &n) != EOF) {        init();        for(int i = 1; i < n; ++i) {            int u, v;            scanf("%d%d", &u, &v);            add(u, v);            add(v, u);        }        dfs(1, -1, 0);        lcaInit();        scanf("%d", &m);        while(m--) {            int u, v;            scanf("%d%d", &u, &v);            int p = lca(u, v);            int dis = dep[u] + dep[v] - 2 * dep[p];            if(dis == 0) {                printf("%d\n", n);                continue;            }            if(dis % 2 == 1) {                puts("0");                continue;            }            if(dep[u] > dep[v]) swap(u, v);            int w = kfa(v, dis / 2);            int x = kfa(v, dis / 2 - 1);            int ans = 0;            if(w == p) {                int y = kfa(u, dis / 2 - 1);                ans = n - sz[x] - sz[y];            }            else {                ans = sz[w] - sz[x];            }            printf("%d\n", ans);        }    }    return 0;}