BestCoder Round #74 (div.2)

组合 1001 LCP Array

第一题就小难，出题的好像是浙大的大牛？

找到一个规律：a[i] = x, s[i..i+x]都想同。a[i] = a[i+1] + 1 (a[i] > 0)，否则就是与后一个颜色不同，方案*25。第一次颜色相同的26种方案。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
int a[N];

int main() {
    int T; scanf ("%d", &T);
    while (T--) {
        int n; scanf ("%d", &n);
        for (int i=1; i<n; ++i) {
            scanf ("%d", a+i);
        }
        a[n] = 0;
        int ans = 26;
        for (int i=n-1; i>=1; --i) {
            if (a[i] == 0) {
                ans = 1ll * ans * 25 % MOD;
            } else if (a[i] != a[i+1] + 1) {
                ans = 0;
                break;
            }
        }
        printf ("%d\n", ans);
    }

    return 0;
}

最短路 1002 Shortest Path

多加了3条边，6个点分别看成起点跑SPFA，然后原来的距离是abs （u - v)，取最小值

#include <bits/stdc++.h>

const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[6], d[6][N], z[N];
bool vis[N];
std::vector<int> G[N];
int n, m;

void add_edge(int u, int v) {
    G[u].push_back (v);
    G[v].push_back (u);
}

void SPFA(int k, int s) {
    memset (d[k], INF, sizeof (d[k]));
    memset (vis, false, sizeof (vis));
    d[k][s] = 0; vis[s] = true;
    std::queue<int> que; que.push (s);
    while (!que.empty ()) {
        int u = que.front (); que.pop ();
        for (auto v: G[u]) {
            if (vis[v]) continue;
            d[k][v] = d[k][u] + 1;
            vis[v] = true;
            que.push (v);
        }
    }
}

int main() {
    int T; scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &m);
        for (int i=1; i<=n; ++i) {
            G[i].clear ();
        }
        for (int i=1; i<n; ++i) {
            add_edge (i, i+1);
        }
        for (int i=0; i<3; ++i) {
            int u, v; scanf ("%d%d", &u, &v);
            add_edge (u, v);
            a[i*2] = u; a[i*2+1] = v;
        }
        for (int i=0; i<6; ++i) {
            SPFA (i, a[i]);
        }
        for (int i=0; i<m; ++i) {
            int u, v; scanf ("%d%d", &u, &v);
            if (u > v) std::swap (u, v);
            int &best = z[i];
            best = v - u;
            for (int i=0; i<6; ++i) {
                best = std::min (best, d[i][u] + d[i][v]);
            }
        }
        int ans = 0;
        for (int i=0; i<m; ++i) {
            ans = (ans + 1ll * (i + 1) * z[i]) % MOD;
        }
        printf ("%d\n", ans);
    }

    return 0;
}

二进制 + BFS 1003 Transform

s -> t = s ^ t = x，所以要预处理出0到x的最小步骤，翻转某一位可以转换为^ (1<<i)

#include <bits/stdc++.h>

const int N = 30;
const int MAX = (1 << 17);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N], d[MAX];
bool vis[N];
int n, m;

void BFS(int s) {
    std::queue<int> que; que.push (s);
    memset (d, INF, sizeof (d)); d[s] = 0;
    while (!que.empty ()) {
        int x = que.front (); que.pop ();
        for (int i=1; i<=n; ++i) {
            int y = x ^ a[i];
            if (y > MAX) break;
            if (d[y] < INF) continue;
            d[y] = d[x] + 1;
            que.push (y);
        }
    }
}

int main() {
    int T; scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &m);
        a[0] = 0;
        for (int i=1; i<=n; ++i) {
            scanf ("%d", a+i);
        }
        int k = 0;
        for (int k=0; k<100; ++k) {
            int x = 1 << k;
            if (x > MAX) break;
            a[++n] = x;
        }
        BFS (0);
        int ans = 0;
        for (int i=0; i<m; ++i) {
            int s, t; scanf ("%d%d", &s, &t);
            int q = s ^ t;
            ans = (ans + 1ll * (i + 1) * d[q]) % MOD;
        }
        printf ("%d\n", ans);
    }

    return 0;
}

1004 Toposort

待补