ZOJ Monthly, June 2014部分题解

感觉挺不错的一套题，写下部分题目的题解

ZOJ 3789： Gears

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3789

题意：有N个齿轮，齿轮间只有旋转方向相反才能连接，给出四个操作

L X Y ：将齿轮X, Y连接，如果它们已经有已属集合，那集合也要合并

D X：拆掉齿轮X，且X所属集合不会断裂

Q X Y：询问X，Y的旋转方向是否相同             

S X：询问X所在集合有多少个齿轮

思路：并查集，维护当前点到根节点的距离及当前点集合的结点个数，方向是否相同可以通过到根节点的距离之差的奇偶性来判断，对于删除结点操作，可以将该点映射到新的结点上，在合并X，Y时要注意更新距离数组

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 600010;int fa[maxn], dis[maxn], num[maxn], p[maxn];void init(){for (int i = 0; i < maxn; i++){fa[i] = i;dis[i] = 0;num[i] = 1;p[i] = i;}}int Find(int x){if (x == fa[x]) return x;int fff = Find(fa[x]);dis[x] += dis[fa[x]];return fa[x] = fff;}int main(){int n, m;while (~scanf("%d%d", &n, &m)){init();int cnt = n;for (int i = 0; i < m; i++){char c[5];scanf("%s", c);if (c[0] == 'L'){int u, v;scanf("%d%d", &u, &v);u = p[u], v = p[v];int fu = Find(u), fv = Find(v);if (fu == fv) continue;fa[fu] = fv;num[fv] += num[fu];dis[fu] = dis[u] + dis[v] + 1;}if (c[0] == 'D'){int u;scanf("%d", &u);int pu = p[u];int fu = Find(pu);num[fu]--;p[u] = ++cnt;}if (c[0] == 'Q'){int u, v;scanf("%d%d", &u, &v);u = p[u], v = p[v];int fu = Find(u), fv = Find(v);if (fu != fv)puts("Unknown");else{if (abs(dis[u] - dis[v]) % 2 == 0)puts("Same");elseputs("Different");}}if (c[0] == 'S'){int u;scanf("%d", &u);u = p[u];printf("%d\n", num[Find(u)]);}}}return 0;}

ZOJ 3790：Consecutive Blocks

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3790

题意：给一个长度为N的序列，你可以从该序列中任意删除K个数，问同一个数字最多能连续多长

思路：先离散化处理，然后对于离散化后的每个数字固定起点，枚举终点进行二分判断，找出最长的结果

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 100010;map <int, int> ma;vector <int> v[maxn];int main(){int n, k;while (~scanf("%d%d", &n, &k)){ma.clear();for (int i = 0; i <= n; i++)v[i].clear();int cnt = 0;for (int i = 0; i < n; i++){int x;scanf("%d", &x);int p = ma[x];if (!p) ma[x] = ++cnt;v[ma[x]].push_back(i);}int ans = 0;for (int i = 1; i <= cnt; i++){int si = v[i].size();for (int j = 0; j < si; j++){int l = -1, r = j;while (r - l > 1){int mid = (l + r) >> 1;int sr = v[i][j], sl = v[i][mid];if (sr - sl - (j - mid) > k)l = mid;elser = mid, ans = max(ans, j - mid + 1);}ans = max(ans, j - r + 1);}}cout << ans << endl;}return 0;}

ZOJ 3791：An Easy Game

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3791

题意：给出两个01字符串s1,s2，每次能且只能改变s1上m个位置的字符，问k次之后有多少种方法让s1变作s2

思路：dp[i][j]表示i次之后两串有j个字符不同，经过一次操作后可以将j个字符中的的t个位置变相同，剩余n - j个位置中m - t个位置变不同，故而有转移方程：dp[i + 1][j - t + m - t] += dp[i][j] * C(j, t) * C(n - j, m - t)，要注意t的枚举范围需满足j - t + m - t <= n

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 200;const int mod = 1e9 + 9;typedef long long ll;ll dp[maxn][maxn], c[maxn][maxn];char s[maxn], ss[maxn];void init(){c[0][0] = 1;for (int i = 1; i <= 100; i++){c[i][0] = 1;for (int j = 1; j <= i; j++)c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;}}int main(){init();int n, m, k;while (~scanf("%d%d%d", &n, &k, &m)){memset(dp, 0, sizeof(dp));scanf("%s", s);scanf("%s", ss);int num = 0;for (int i = 0; i < n; i++)if (s[i] != ss[i]) num++;dp[0][num] = 1;for (int i = 0; i < k; i++)for (int j = 0; j <= n; j++){if (dp[i][j] == 0) continue;for (int t = max((j + m - n) / 2, 0); t <= j && t <= m; t++)dp[i + 1][j - t + m - t] = (dp[i + 1][j - t + m - t] + dp[i][j] * c[j][t] % mod * c[n - j][m - t] % mod) % mod;}cout << dp[k][0] << endl;}return 0;}

ZOJ 3792：Romantic Value

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3792

题意：给一张无向图，给出起点和终点，要删除一些边使起点到终点不连通，在删除边权值最小的情况下保证删除边的数目最少，求剩余边权值和 / 删除边数的这一比值

思路：可以构造每条边的边权为w *  maxn + 1，这样就不会出现多种最小割的方案了，注意maxn要较大使得边数不会影响网络流的流向，最终跑出来的结果ans / maxn为最小割，ans % maxn为删除边数

也可以在原图跑完最大流的基础上，找出那些满流的边将其流量改为1，其余边流量改为INF，再跑一次最大流也可以找出删除的最少边数

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int MAXN = 1010;const int MAXM = 50010;const int INF = 0x3f3f3f3f;typedef long long ll;struct Edge{int to, next, cap, flow;} edge[MAXM];int tol;int head[MAXN];int gap[MAXN], dep[MAXN], cur[MAXN];void init(){tol = 0;memset(head, -1, sizeof(head));}void addedge(int u, int v, int w, int rw = 0){edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++;}int Q[MAXN];void BFS(int start, int end){memset(dep, -1, sizeof(dep));memset(gap, 0, sizeof(gap));gap[0] = 1;int front = 0, rear = 0;dep[end] = 0;Q[rear++] = end;while (front != rear){int u = Q[front++];for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if (dep[v] != -1)continue;Q[rear++] = v;dep[v] = dep[u] + 1;gap[dep[v]]++;}}}int S[MAXN];int sap(int start, int end, int N){BFS(start, end);memcpy(cur, head, sizeof(head));int top = 0;int u = start;int ans = 0;while (dep[start] < N){if (u == end){int Min = INF;int inser;for (int i = 0; i < top; i++)if (Min > edge[S[i]].cap - edge[S[i]].flow){Min = edge[S[i]].cap - edge[S[i]].flow;inser = i;}for (int i = 0; i < top; i++){edge[S[i]].flow += Min;edge[S[i] ^ 1].flow -= Min;}ans += Min;top = inser;u = edge[S[top] ^ 1].to;continue;}bool flag = false;int v;for (int i = cur[u]; i != -1; i = edge[i].next){v = edge[i].to;if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]){flag = true;cur[u] = i;break;}}if (flag){S[top++] = cur[u];u = v;continue;}int Min = N;for (int i = head[u]; i != -1; i = edge[i].next)if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){Min = dep[edge[i].to];cur[u] = i;}gap[dep[u]]--;if (!gap[dep[u]])return ans;dep[u] = Min + 1;gap[dep[u]]++;if (u != start)u = edge[S[--top] ^ 1].to;}return ans;}int main(){int tt;cin >> tt;while (tt--){init();int n, m, p, q;cin >> n >> m >> p >> q;int sum = 0;for (int i = 0; i < m; i++){int u, v, w;scanf("%d%d%d", &u, &v, &w);sum += w;addedge(u, v, w * MAXN + 1);addedge(v, u, w * MAXN + 1);}int ans = sap(p, q, n + 1);if (ans == 0)puts("Inf");elseprintf("%.2f\n", 1.0 * (sum - ans / MAXN) / (ans % MAXN));}return 0;}

ZOJ 3793：First Digit

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3793

题意：本福特定律的介绍

思路：数字1是满足那个概率范围的，所以输出1即可

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 100010;typedef long long ll;int a[maxn];int main(){int t;cin >> t;while (t--){int b, e;cin >> b >> e;cout << 1 << endl;}return 0;}

ZOJ  3794：Greedy Driver

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3794

题意：一个司机要从1走到N，他可以在任意有加油站的地点加油，走每条道路需要消耗相应的油量，有些城市可以通过卖油赚钱，但只能卖一次油，如果司机不能走到N，输出-1，如果能走到输出最多能赚多少钱

思路：从起点跑一次spfa算出到每个点最多能剩多少油，再从起点反向跑一次spfa算出每个点到终点的最短距离，即最少需要多少油到终点，然后枚举每个可以卖油的城市更新答案，在跑spfa的过程中要注意对加油站的处理

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 1010;const int maxm = 100010;const int inf = 0x3f3f3f3f;int n, m, c;int cnt, head[maxn];int f[maxn], vis[maxn], dis[maxn];int p[maxn], q[maxn];struct edge{int to, w, nxt, id;} e[maxm << 1];void init(){cnt = 0;memset(head, -1, sizeof(head));}void add(int u, int v, int w, int id){e[cnt].to = v;e[cnt].w = w;e[cnt].id = id;e[cnt].nxt = head[u];head[u] = cnt++;}void spfa(int s){memset(vis, 0, sizeof(vis));memset(f, -1, sizeof(f));queue <int> que;que.push(s);f[s] = c, vis[s] = 1;while (!que.empty()){int u = que.front();que.pop();vis[u] = 0;for (int i = head[u]; ~i; i = e[i].nxt){if (e[i].id == 1) continue;int v = e[i].to;if (f[u] >= e[i].w){int tmp = f[u] - e[i].w;if (p[v]) tmp = c;if (tmp > f[v]){f[v] = tmp;if (!vis[v]){vis[v] = 1;que.push(v);}}}}}}void rspfa(int s){memset(vis, 0, sizeof(vis));memset(dis, inf, sizeof(dis));queue <int> que;que.push(s);dis[s] = 0, vis[s] = 1;while (!que.empty()){int u = que.front();que.pop();vis[u] = 0;for (int i = head[u]; ~i; i = e[i].nxt){if (e[i].id == 0) continue;int v = e[i].to;if (dis[v] > dis[u] + e[i].w){dis[v] = dis[u] + e[i].w;if (p[v]) dis[v] = 0;if (!vis[v]){vis[v] = 1;que.push(v);}}}}}int main(){while (~scanf("%d%d%d", &n, &m, &c)){init();for (int i = 0; i < m; i++){int u, v, w;scanf("%d%d%d", &u, &v, &w);add(u, v, w, 0);add(v, u, w, 1);}memset(p, 0, sizeof(p));memset(q, 0, sizeof(q));int P, Q;scanf("%d", &P);while (P--){int x;scanf("%d", &x);p[x] = 1;}scanf("%d", &Q);while (Q--){int x, v;scanf("%d%d", &x, &v);q[x] = v;}spfa(1);if (f[n] == -1){puts("-1");continue;}rspfa(n);// for (int i = 1; i <= n; i++)// printf("************%d %d\n", f[i], dis[i]);int ans = 0;for (int i = 1; i <= n; i++){if (q[i] && f[i] >= dis[i])ans = max(ans, (f[i] - dis[i]) * q[i]);}cout << ans << endl;}return 0;}

ZOJ 3795：Grouping

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3795

题意：给定N个点M条单向边，问最少需要分多少个集合，使得集合内的点都不能达到其他点

思路：对于强连通分量内的点肯定不能在一个集合，另外该分量所在的那一条路径上的点也不能在同一个集合，故缩点后处理出一条点权和最大的路径即是答案

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 101000;const int maxm = 300010;const int inf = 0x3f3f3f3f;int cnt, head[maxn];int low[maxn], dfn[maxn], sta[maxn], bel[maxn];int idx, top, scc;bool insta[maxn];int num[maxn];struct edge{int to, w, nxt;} e[maxm];void init(){cnt = 0;memset(head, -1, sizeof(head));}void add(int u, int v){e[cnt].to = v;e[cnt].nxt = head[u];head[u] = cnt++;}void tarjan(int u){int v;low[u] = dfn[u] = ++idx;sta[top++] = u;insta[u] = true;for (int i = head[u]; ~i; i = e[i].nxt){int v = e[i].to;if (!dfn[v]){tarjan(v);if (low[u] > low[v])low[u] = low[v];}else if (insta[v] && low[u] > dfn[v])low[u] = dfn[v];}if (low[u] == dfn[u]){scc++;do{v = sta[--top];insta[v] = false;bel[v] = scc;num[scc]++;} while (v != u);}}void solve(int n){idx = scc = top = 0;memset(dfn, 0, sizeof(dfn));memset(insta, 0, sizeof(insta));memset(num, 0, sizeof(num));for (int i = 1; i <= n; i++)if (!dfn[i])tarjan(i);}vector <int> g[maxn];int dp[maxn];int dfs(int u){if (dp[u] != -1) return dp[u];int res = 0;for (int i = 0; i < g[u].size(); i++){int v = g[u][i];res = max(res, dfs(v));}dp[u] = res + num[u];return dp[u];}int main(){int n, m;while (~scanf("%d%d", &n, &m)){init();for (int i = 0; i < m; i++){int u, v;scanf("%d%d", &u, &v);add(u, v);}solve(n);memset(dp, -1, sizeof(dp));for (int i = 0; i <= scc; i++)g[i].clear();for (int u = 1; u <= n; u++){for (int i = head[u]; ~i; i = e[i].nxt){int v = e[i].to;if (bel[u] != bel[v])g[bel[u]].push_back(bel[v]);}}// for (int i = 1; i <= scc; i++)// for (int j = 0; j < g[i].size(); j++)// printf("********** %d\n", g[i][j]);int ans = 0;for (int i = 1; i <= scc; i++)ans = max(ans, dfs(i));cout << ans << endl;}return 0;}