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9.6线上模拟

（）

T1 纯纯模拟但是超时了（）极度怀疑是数据类型的锅：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <climits>
 5 #include <cassert>
 6 #include <cmath>
 7 #include <cstring>
 8 #include <cstdlib>
 9 #include <cctype>
10 #include <utility>
11 #include <set>
12 #include <map>
13 #include <stack>
14 #include <queue>
15 #include <vector>
16 #define int long long
17 #define ll long long
18 #define ull unsigned long long
19 #define re register
20 #define lowbit(x) x & (-x)
21 #define gcd(a,b) _gcd(a,b)
22 #define mid ((l + r) >> 1)
23 #define rep(i,a,b)  for(re int i(a);i <= b;i ++)
24 #define rrep(i,a,b) for(re int i(a);i >= b;i --)
25 using namespace std;
26 inline int read(){
27     re int x = 0,f = 1;
28     re char ch = getchar();
29     while(ch < '0' || ch > '9'){
30         if(ch == '-') f = -1;
31         ch = getchar();
32     }
33     while(ch >= '0' && ch <= '9'){
34         x = (x << 3) + (x << 1) + (ch ^ 48);
35         ch = getchar();
36     }
37     return x * f;
38 }
39 const int M = 1e5 + 1;
40 struct A{
41     double a,b;
42 }a[M];
43 double ans = 0.0;
44 int n;
45 signed main(){
46         freopen("money.in","r",stdin);
47         freopen("money.out","w",stdout);
48         n = read();
49         rep(i,1,n){
50             scanf("%lf%lf",&a[i].a,&a[i].b);
51             ans += a[i].a * a[i].b / 10;
52         }
53     printf("%.2lf\n",ans);
54     return 0;
55 }

T2 待更。。。。。

T3 半联通

由于单向（定义）

所以最大的半连通子图肯定是一条链

那就先缩点点权为分量顶点数然后去重边

f[i]表示到第i的最长路径 ans[i]表示到第i个点的最长路径数量 rcd[]计入度

拓扑序DP跑最长链

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <climits>
  5 #include <cassert>
  6 #include <cmath>
  7 #include <cstring>
  8 #include <cstdlib>
  9 #include <cctype>
 10 #include <utility>
 11 #include <set>
 12 #include <map>
 13 #include <queue>
 14 #include <vector>
 15 #define int long long
 16 #define ll long long
 17 #define ull unsigned long long
 18 #define re register
 19 #define lowbit(x) x & (-x)
 20 #define gcd(a,b) _gcd(a,b)
 21 #define mid ((l + r) >> 1)
 22 #define rep(i,a,b)  for(re int i(a);i <=  b;i ++)
 23 #define rrep(i,a,b) for(re int i(a);i >=  b;i --)
 24 using namespace std;
 25 inline int read(){
 26     re  int x  =  0,f  =  1;
 27     re  char ch  =  getchar();
 28     while(ch < '0' || ch > '9'){
 29         if(ch  ==  '-') f  =  -1;
 30         ch  =  getchar();
 31     }
 32     while(ch >=  '0' && ch <=  '9'){
 33         x  =  (x << 3) + (x << 1) + (ch ^ 48);
 34         ch  =  getchar();
 35     }
 36     return x * f;
 37 }
 38 const int M = 2000010;
 39 int n,m,p;
 40 int cnt  =  0;
 41 int top;
 42 int tmp;
 43 int head[M];
 44 int tot;
 45 struct node{
 46     int s;
 47     int to;
 48     int next;
 49 }b[M];
 50 node e[M];
 51 struct A{
 52     int x;
 53     int y;
 54 }a[M];
 55 int head1[M];
 56 int tot1  =  0;
 57 int rcd[M];
 58 int ans[M];
 59 int f[M];
 60 int c[M];
 61 int st[M];
 62 int low[M];
 63 int dfn[M];
 64 int d[M];
 65 bool vis[M];
 66 queue<int> q;
 67 inline void add(int x,int y){b[++ tot] = (node){x,y,head[x]};head[x] = tot;}
 68 inline void add1(int x,int y){e[++ tot1] = (node){x,y,head1[x]};head1[x] = tot1;}
 69 inline void tarjan(int x){
 70     low[x] = dfn[x] = ++ cnt;
 71     st[++ top] = x;
 72     vis[x] = 1;
 73     for(re int i(head[x]);i;i = b[i].next){
 74         int y = b[i].to;
 75         if(!dfn[y]){
 76             tarjan(y);
 77             low[x] = min(low[x],low[y]);
 78         }
 79         else if(vis[y]) low[x] = min(low[x], dfn[y]);
 80     }
 81     if(low[x] == dfn[x]){
 82         tmp ++;
 83         do{
 84             c[st[top]] = tmp;
 85             d[tmp] ++;
 86             vis[st[top]] = 0;
 87             top --;
 88         }
 89         while(st[top + 1]!= x);
 90     }
 91     return;
 92 }
 93 int maxx = 0;
 94 inline bool cmp(A x, A y){return x.x == y.x ? x.y < y.y : x.x < y.x;}
 95 signed main(){
 96     freopen("semi.in","r",stdin);
 97     freopen("semi.out","w",stdout);
 98     n = read();
 99     m = read();
100     p = read();
101     rep(i,1,m){
102         a[i].x = read();
103         a[i].y = read();
104         add(a[i].x,a[i].y);
105     }
106     rep(i,1,n) if(!dfn[i]) tarjan(i);
107     rep(i,1,m) a[i].x = c[a[i].x],a[i].y = c[a[i].y];
108     sort(a + 1,a + 1 + m,cmp);
109     rep(i,1,m) if(a[i].x!= a[i].y && (a[i].x != a[i-1].x || a[i].y != a[i-1].y)) add1(a[i].x, a[i].y), rcd[a[i].y] ++;
110     rep(i,1,tmp) {
111         ans[i] = 1,f[i] = d[i];
112         if (!rcd[i]) q.push(i);
113     }
114     while(!q.empty()){
115         int x = q.front();
116         q.pop();
117         maxx = max(maxx, f[x]);
118         for(re int i(head1[x]);i;i = e[i].next){
119             int y = e[i].to;
120             rcd[y] --;
121             if(f[y] < f[x] + d[y]){
122                 f[y] = f[x] + d[y];
123                 ans[y] = ans[x];
124             }
125             else if(f[y] == f[x]+d[y]) ans[y] = (ans[y] + ans[x]) % p;
126             if(!rcd[y]) q.push(y);
127         }
128     }
129     cnt = 0;
130     rep(i,1,tmp) if(maxx == f[i]) cnt = (cnt + ans[i]) % p;
131     printf("%d\n%d",maxx,cnt % p);
132     return 0;
133 }