「网络流 24 题」最长递增子序列

1：dp求一遍

2：所有点拆成入点和出点，对于没个f[i]=1的1点连源点，每个f[i]=ans1的点连汇点，每个点的入点和出点再连一条边，所有容量为1，求一遍最大流。

2：1和n可以用多次，所以对于点1和点n，入点和出的连边的容量变为INF，如果需要连源点或汇点那么容量也变为n，其余容量为1，求一遍最大流。

#include<stdio.h>#include<algorithm>#include<string.h>#include<queue>using namespace std;const int maxm = 10005;const int maxn = 1000005;const int INF = 1e9 + 7;struct node{int v, flow, next;}edge[maxm];int dis[maxm], dep[maxm], cur[maxm], head[maxm], f[maxm], vis[maxm], pre[maxm], a[maxm];int n, m, s, t, cnt;void init(){cnt = 0, s = 0, t = n * 2 + 1;memset(head, -1, sizeof(head));}void add(int u, int v, int w){edge[cnt].v = v, edge[cnt].flow = w, edge[cnt].next = head[u], head[u] = cnt++;edge[cnt].v = u, edge[cnt].flow = 0, edge[cnt].next = head[v], head[v] = cnt++;}int bfs(){queue<int>q;memset(dis, -1, sizeof(dis));memset(pre, -1, sizeof(pre));dis[s] = 0;q.push(s);while (!q.empty()){int u = q.front();q.pop();for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] == -1 && edge[i].flow){dis[v] = dis[u] + 1;pre[v] = u;q.push(v);}}}if (dis[t] == -1) return 0;return 1;}int dfs(int u, int flow){if (u == t)return flow;for (int i = cur[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] == dis[u] + 1 && edge[i].flow){int d = dfs(v, min(flow, edge[i].flow));if (d > 0){edge[i].flow -= d, edge[i ^ 1].flow += d;return d;}}}return 0;}int dinic(){int ans = 0, d;while (bfs()){for (int i = s;i <= t;i++) cur[i] = head[i];while (d = dfs(s, INF))ans += d;}return ans;}int main(){int i, j, k, sum, ans = 1;scanf("%d", &n);init();for (i = 1;i <= n;i++){scanf("%d", &a[i]);f[i] = 1;for (j = 1;j < i;j++)if (a[j] <= a[i]){f[i] = max(f[i], f[j] + 1);ans = max(ans, f[i]);}}printf("%d\n", ans);for (i = 1;i <= n;i++){if (f[i] == 1) add(s, i, 1);if (f[i] == ans) add(i + n, t, 1);add(i, i + n, 1);for (j = 1;j < i;j++)if (f[i] == f[j] + 1 && a[i] >= a[j])add(j + n, i, 1);}if (ans == 1) printf("%d\n", n);else printf("%d\n", dinic());init();for (i = 1;i <= n;i++){int xx = 1;if (i == 1 || i == n) xx = INF;if (f[i] == 1) add(s, i, xx);if (f[i] == ans) add(i + n, t, xx);add(i, i + n, xx);for (j = 1;j < i;j++)if (f[i] == f[j] + 1 && a[i] >= a[j])add(j + n, i, 1);}if (ans == 1) printf("%d\n", n);else printf("%d\n", dinic());return 0;}