[20151015]SCZ训练

T1：

在一个有N(1≤N≤10000)个点，M(1≤M≤50000)条边的图中，给出K个特殊点（1≤K≤5），要求从剩下的点中选一个点作为起点，至少经过每一个特殊点一次后回到起点，求最短距离。

Solution ：
K!枚举经过点的顺序，预处理出特殊点与所有点的最短距离，然后O(N)枚举起点即可

Code：

/*************************************************************************    > File Name: relocation.cpp    > Author: Archer ************************************************************************/#include <bits/stdc++.h>#include <ext/pb_ds/priority_queue.hpp>#include <ext/pb_ds/hash_policy.hpp>using namespace std;const int N = 11111;const int M = 55555;const int K = 10;const int INF = 2147483647;#define REP(i, l, r) for (int i = l; i <= r; i++)struct E{ int to, c, nxt; }edge[M << 1];int top = 0, last[N];bool vis[N], shut[N];int a[K], ans, dist[K][N], q[N << 2], n, m, k;inline void setIO(){    freopen("relocation.in", "r", stdin);    freopen("relocation.out", "w", stdout);}inline int read(){    int x = 0, f = 1; char ch = getchar();    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar(); }    return x * f;}inline void addedge(int u, int v, int c){    edge[++top].to = v; edge[top].c = c; edge[top].nxt = last[u]; last[u] = top;    edge[++top].to = u; edge[top].c = c; edge[top].nxt = last[v]; last[v] = top;}inline void SPFA(int S, int no){    int l, r, now; memset(vis, 0, sizeof(vis));    REP(i, 1, n) dist[no][i] = INF;    for (dist[no][q[l = r = 1] = S] = 0; l <= r; l++){        now = q[l]; vis[now] = 0;        for (int p = last[now]; p; p = edge[p].nxt)            if (dist[no][edge[p].to] > dist[no][now] + edge[p].c){                dist[no][edge[p].to] = dist[no][now] + edge[p].c;                if (!vis[edge[p].to]) vis[q[++r] = edge[p].to] = 1;            }    }}inline void check(int S){    int res = 0;    REP(i, 1, k - 1) res += dist[q[i]][a[q[i + 1]]];    ans = min(ans, res + dist[q[1]][S] + dist[q[k]][S]);}inline void dfs(int dep, int S){    if (dep > k){check(S); return;}    REP(i, 1, k)        if (!vis[i]){            vis[i] = 1; q[dep] = i;            dfs(dep + 1, S);            vis[i] = 0;        }}int main(){    setIO();    n = read(); m = read(); k = read();    REP(i, 1, k) shut[a[i] = read()] = 1;    REP(i, 1, m){        int u = read(), v = read(), c = read();        addedge(u, v, c);    }    REP(i, 1, k) SPFA(a[i], i);    ans = INF; memset(vis, 0, sizeof(vis));    REP(i, 1, n) if (!shut[i]) dfs(1, i);    printf("%d\n", ans);    return 0;}

T2 :

招募n个男的m个女的，每个人起初需要10000元，这些男女中存在朋友关系，假如在招募某个人之前已经招募了与这个人有朋友关系的人，招募费用就可以减少，减少的量为亲密度。求最少招募费用。

Solution :
显然对于一种招募顺序，有一棵生成树与之对应，我们求亲密度最大生成树，用(n+m)∗10000−W即为答案

Code :

/*************************************************************************    > File Name: conscription.cpp    > Author: Archer ************************************************************************/#include <bits/stdc++.h>#include <ext/pb_ds/priority_queue.hpp>#include <ext/pb_ds/hash_policy.hpp>using namespace std;const int N = 22222;const int M = 55555;#define REP(i, l, r) for (int i = l; i <= r; i++)struct E{int u, v, c;}edge[M << 1];int m, n, r, fa[M << 1];inline void setIO(){    freopen("conscription.in", "r", stdin);    freopen("conscription.out", "w", stdout);}inline int read(){    int x = 0, f = 1; char ch = getchar();    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar(); }    return x * f;}inline bool cmp(E e1, E e2){ return e1.c > e2.c; }inline int find(int k){ return fa[k] = fa[k] == k ? k : find(fa[k]); }int main(){    setIO();    m = read(); n = read(); r = read();    REP(i, 1, r)         edge[i].u = read() + 1, edge[i].v = read() + 1 + m, edge[i].c = read();    REP(i, r + 1, r + n + m)        edge[i].u = i, edge[i].v = n + m + 1, edge[i].c = 0;    sort(edge + 1, edge + r + n + m, cmp);    REP(i, 1, n + m + 1) fa[i] = i;    int ans = 0, cnt = 0;    REP(i, 1, r + n + m){        if (find(edge[i].u) == find(edge[i].v)) continue;        cnt++;        fa[find(edge[i].u)] = find(edge[i].v);        ans += edge[i].c;        if (cnt == n + m) break;    }    printf("%d\n", (n + m) * 10000 - ans);    return 0;}

T3 :

求[0,m−1](1≤m≤231−1)之间与Z(Z≤m)有至少一个起始位置相同且长度为r(r≤18)的公共子串的数的个数

Solution :
数位DP。记dp[i][j][less]表示当前匹配到第i位，已经连续匹配了j个字符，是否卡死在上限的方案数，dfs转移。

Code :

/*************************************************************************    > File Name: ticket.cpp    > Author: Archer ************************************************************************/#include <bits/stdc++.h>#include <ext/pb_ds/priority_queue.hpp>#include <ext/pb_ds/hash_policy.hpp>using namespace std;typedef long long ll;typedef unsigned long long ull;const int N = 111;ll dp[N][N][2], m, z, r, lenm, lenz;char stm[N], stz[N];inline void setIO(){    freopen("ticket.in", "r", stdin);    freopen("ticket.out", "w", stdout);}inline ll read(){    ll x = 0ll, f = 1ll; char ch = getchar();    while (!isdigit(ch)) {if (ch == '-') f = -1ll; ch = getchar();}    while (isdigit(ch)) {x = x * 10ll + 1ll*(ch - '0'); ch = getchar(); }    return x * f;}inline ll dfs(bool flag, int dep, int cnt){    ll res = 0;    if (~dp[dep][cnt][flag]) return dp[dep][cnt][flag];    if (cnt == r){        if (flag) return dp[dep][cnt][flag] = pow(10, lenm - dep);        long long res = 0;        for (int i = dep; i < lenm; i++)                res = res * 10 + stm[i] - '0';        return dp[dep][cnt][flag] = res + 1;    }    if (dep >= lenm) return 0;    for (char c = '0'; c <= (flag ? '9' : stm[dep]); ++c){        res += dfs(flag || c < stm[dep], dep + 1, c == stz[dep] ? cnt + 1 : 0);    }    return dp[dep][cnt][flag] = res;}int main(){    setIO();    int case_cnt = read();    while (case_cnt--){        m = read() - 1; z = read(); r = read();         lenm = 0; lenz = 0;        while (m > 0){ stm[lenm++] = (char)(m % 10ll + '0'); m /= 10ll; }        while (z > 0){ stz[lenz++] = (char)(z % 10ll + '0'); z /= 10ll; }        for (int i = lenz; i < lenm; i++) stz[i] = '0';        for (int i = 0; i < lenm / 2; i++) swap(stm[i], stm[lenm - i - 1]);        for (int i = 0; i < lenm / 2; i++) swap(stz[i], stz[lenm - i - 1]);        memset(dp, 0xff, sizeof(dp));        printf("%I64d\n", dfs(0, 0, 0));        }    return 0;}