kyeremal-网络流24题T6-最长递增子序列问题

来源：互联网发布：excel表格查找重复数据编辑：程序博客网时间：2024/06/05 18:37

原题：

共三问：

（1）最长上升子序列长度s

（2）每个数用一遍，能够得到的最多长度为s的子序列个数

（3）同第二问，a[1]与a[n]可用多次

分析：

（1）对于第一问，直接dp。

（2）第二问，拆点后建图，根据第一问求出的LIS，如果LIS[i] = LIS[j]+1 && i < j && a[i] < a[j]，则连接一条i->j容量为1的有向边，最后左一边最大流即可。

（3）第三问，同第二问建图方法，同时S->1, n->T的流量为正无穷。

code：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define rep(i, l, r) for (int i = l; i <= r; i++)#define REP(i, l, r) for (int i = l; i >= r; i--)#define INF 19971228#define MAXN 1010int n, N = -1, seq[MAXN], f[MAXN], maxs = -INF, S, T, first[MAXN], next[MAXN], dis[MAXN];struct tlist {int x, y, f;} a[MAXN];queue<int> q;inline int min(int a, int b) {return a<b ? a : b;}inline int max(int a, int b) {return a>b ? a : b;}inline void add(int x, int y, int f) {a[++N].x = x, a[N].y = y, a[N].f = f, next[N] = first[x], first[x] = N;a[++N].x = y, a[N].y = x, a[N].f = 0, next[N] = first[y], first[y] = N;}inline bool bfs() {while (!q.empty()) q.pop();memset(dis, -1, sizeof(dis));q.push(S);dis[S] = 0;while (!q.empty()) {int x = q.front();q.pop();for (int i = first[x]; ~i; i = next[i])if (!~dis[a[i].y] && a[i].f) {dis[a[i].y] = dis[x] + 1;q.push(a[i].y);}}return ~dis[T];}inline int find(int x, int low) {if (x == T) return low;int sum = 0, temp;for (int i = first[x]; ~i; i = next[i])if ((a[i].f) && (dis[a[i].y] == dis[x] + 1) && (temp = find(a[i].y, min(low-sum, a[i].f)))) {a[i].f -= temp;a[i^1].f += temp;sum += temp;}return sum;}inline int dinic() {int ans = 0, temp;while (bfs())while (temp = find(S, 0x7fffffff))ans += temp;return ans;}int main() {cin >> n;S = 0, T = 2*n+1;memset(first, -1, sizeof(first));memset(next, -1, sizeof(next));rep(i, 1, n) scanf("%d", &seq[i]);rep(i, 0, n) f[i] = 1;REP(i, n-1, 1) {int cur = f[i];rep(j, i+1, n) if (seq[j] > seq[i]) f[i] = max(f[i], cur + f[j]);}rep(i, 1, n) maxs = max(maxs, f[i]);cout << maxs << endl;rep(i, 1, n) {if (f[i] == maxs) add(S, i, 1);if (f[i] == 1) add(n+i, T, 1);}rep(i, 1, n) add(i, n+i, 1);rep(i, 1, n)rep(j, i+1, n)if ((seq[j] > seq[i]) && (f[i] == f[j]+1)) add(n+i, j, 1);cout << dinic() << endl;N = -1;memset(first, -1, sizeof(first));memset(next, -1, sizeof(next));if (f[1] == maxs) add(S, 1, INF);if (f[1] == 1) add(n+1, T, 1);rep(i, 2, n-1) {if (f[i] == maxs) add(S, i, 1);if (f[i] == 1) add(n+i, T, 1);}if (f[n] == maxs) add(S, n, 1);if (f[n] == 1) add(n+n, T, INF);add(1, n+1, INF);rep(i, 2, n-1) add(i, n+i, 1);add(n, n+n, INF);rep(i, 1, n)rep(j, i+1, n)if ((seq[j] > seq[i]) && (f[i] == f[j]+1))add(n+i, j, 1);cout << dinic() << endl;return 0;}

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