[swust 1741] 最长递增子序列问题（DP，最大流）

题目链接：https://www.oj.swust.edu.cn/problem/show/1741

此题同上，但是多一个问：x1和xn能用多次。

解法一样，只不过这两个点相关的边都是inf就行了。这样可以表示能无限用。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3
  4 typedef struct Edge {
  5     int u, v, w, next;
  6 }Edge;
  7
  8 const int inf = 0x7f7f7f7f;
  9 const int maxn = 6660;
 10
 11 int cnt, dhead[maxn];
 12 int cur[maxn], dd[maxn];
 13 Edge dedge[maxn<<8];
 14 int S, T, N;
 15
 16 void init() {
 17     memset(dhead, -1, sizeof(dhead));
 18     for(int i = 0; i < maxn; i++) dedge[i].next = -1;
 19     S = 0; cnt = 0;
 20 }
 21
 22 void adde(int u, int v, int w, int c1=0) {
 23     dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w;
 24     dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
 25     dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1;
 26     dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
 27 }
 28
 29 bool bfs(int s, int t, int n) {
 30     queue<int> q;
 31     for(int i = 0; i < n; i++) dd[i] = inf;
 32     dd[s] = 0;
 33     q.push(s);
 34     while(!q.empty()) {
 35         int u = q.front(); q.pop();
 36         for(int i = dhead[u]; ~i; i = dedge[i].next) {
 37             if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
 38                 dd[dedge[i].v] = dd[u] + 1;
 39                 if(dedge[i].v == t) return 1;
 40                 q.push(dedge[i].v);
 41             }
 42         }
 43     }
 44     return 0;
 45 }
 46
 47 int dinic(int s, int t, int n) {
 48     int st[maxn], top;
 49     int u;
 50     int flow = 0;
 51     while(bfs(s, t, n)) {
 52         for(int i = 0; i < n; i++) cur[i] = dhead[i];
 53         u = s; top = 0;
 54         while(cur[s] != -1) {
 55             if(u == t) {
 56                 int tp = inf;
 57                 for(int i = top - 1; i >= 0; i--) {
 58                     tp = min(tp, dedge[st[i]].w);
 59                 }
 60                 flow += tp;
 61                 for(int i = top - 1; i >= 0; i--) {
 62                     dedge[st[i]].w -= tp;
 63                     dedge[st[i] ^ 1].w += tp;
 64                     if(dedge[st[i]].w == 0) top = i;
 65                 }
 66                 u = dedge[st[top]].u;
 67             }
 68             else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
 69                 st[top++] = cur[u];
 70                 u = dedge[cur[u]].v;
 71             }
 72             else {
 73                 while(u != s && cur[u] == -1) {
 74                     u = dedge[st[--top]].u;
 75                 }
 76                 cur[u] = dedge[cur[u]].next;
 77             }
 78         }
 79     }
 80     return flow;
 81 }
 82
 83 int n;
 84 int x[maxn];
 85 int dp[maxn];
 86 int ret;
 87
 88 void work1() {
 89     ret = 0;
 90     memset(dp, 0, sizeof(dp));
 91     for(int i = 1; i <= n; i++) {
 92         dp[i] = 1;
 93         for(int j = 1; j <= i; j++) {
 94             if(x[i] > x[j]) dp[i] = max(dp[i], dp[j]+1);
 95         }
 96         ret = max(ret, dp[i]);
 97     }
 98     printf("%d\n", ret);
 99 }
100
101 void work2() {
102     if(ret == 1) {
103         printf("%d\n", n);
104         return;
105     }
106     init();
107     S = 0, T = 2 * n + 1, N = T + 1;
108     for(int i = 1; i <= n; i++) {
109         adde(i, i+n, 1);
110         if(dp[i] == 1) adde(S, i, 1);
111         if(dp[i] == ret) adde(i+n, T, 1);
112     }
113     for(int i = 2; i <= n; i++) {
114         for(int j = 1; j < i; j++) {
115             if(x[i] > x[j] && dp[i] == dp[j] + 1) {
116                 adde(j+n, i, 1);
117             }
118         }
119     }
120     cout << dinic(S, T, N) << endl;
121 }
122
123 void work3() {
124     if(ret == 1) {
125         printf("%d\n", n);
126         return;
127     }
128     init();
129     S = 0, T = 2 * n + 1, N = T + 1;
130     for(int i = 1; i <= n; i++) {
131         if(i == 1 || i == n) {
132             adde(i, i+n, inf);
133             if(dp[i] == 1) adde(S, i, inf);
134             if(dp[i] == ret) adde(i+n, T, inf);
135         }
136         else {
137             adde(i, i+n, 1);
138             if(dp[i] == 1) adde(S, i, 1);
139             if(dp[i] == ret) adde(i+n, T, 1);
140         }
141     }
142     for(int i = 2; i <= n; i++) {
143         for(int j = 1; j < i; j++) {
144             if(x[i] > x[j] && dp[i] == dp[j] + 1) {
145                 adde(j+n, i, 1);
146             }
147         }
148     }
149     cout << dinic(S, T, N) << endl;
150 }
151
152 int main() {
153     // freopen("in", "r", stdin);
154     while(~scanf("%d",&n)) {
155         init();
156         for(int i = 1; i <= n; i++) scanf("%d", &x[i]);
157         work1(); work2();
158     }
159     return 0;
160 }