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 ", ret); 99 } 100 101 void work2() { 102 if(ret == 1) { 103 printf("%d ", 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); 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 }