输入输出样例
输入 #1
9 2 4 1 3 5 6 3 3 5 6
输出 #1
2
输入 #2
9 3 4 1 3 5 6 3 3 5 6 3 1 5 9
输出 #2
3
思路
这题想了将近一晚上，从DAG上最长路想到暴搜想到差分约束，多次无果决定看看算法标签。。
拓扑序。。
然后想到可以把时间序转化成等级序来做。
然而wa了一个点，明早再重构写一次吧。
UPD : 已更新代码。。。
CODE
1 #include <bits/stdc++.h> 2 #define dbg(x) cout << #x << "=" << x << endl 3 #define eps 1e-8 4 #define pi acos(-1.0) 5 6 using namespace std; 7 typedef long long LL; 8 9 template<class T>inline void read(T &res) 10 { 11 char c;T flag=1; 12 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 13 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 14 } 15 16 namespace _buff { 17 const size_t BUFF = 1 << 19; 18 char ibuf[BUFF], *ib = ibuf, *ie = ibuf; 19 char getc() { 20 if (ib == ie) { 21 ib = ibuf; 22 ie = ibuf + fread(ibuf, 1, BUFF, stdin); 23 } 24 return ib == ie ? -1 : *ib++; 25 } 26 } 27 28 int qread() { 29 using namespace _buff; 30 int ret = 0; 31 bool pos = true; 32 char c = getc(); 33 for (; (c < '0' || c > '9') && c != '-'; c = getc()) { 34 assert(~c); 35 } 36 if (c == '-') { 37 pos = false; 38 c = getc(); 39 } 40 for (; c >= '0' && c <= '9'; c = getc()) { 41 ret = (ret << 3) + (ret << 1) + (c ^ 48); 42 } 43 return pos ? ret : -ret; 44 } 45 46 const int maxn = 1e3 + 7; 47 const int mmm = 5e6 + 7; 48 49 int in[maxn]; 50 int edge[mmm], head[mmm], nxt[mmm], cnt; 51 int n, m, ans; 52 53 int a[maxn]; 54 bool vis[maxn][maxn]; 55 int depth[maxn]; 56 int stop[maxn]; 57 58 void BuildGraph(int u, int v) { 59 cnt++; 60 edge[cnt] = v; 61 nxt[cnt] = head[u]; 62 head[u] = cnt; 63 } 64 65 void topo() { 66 queue <int> q; 67 for ( int i = 1; i <= n; ++i ) { 68 if (in[i] == 0) { 69 q.push(i); 70 depth[i] = 1; 71 } 72 } 73 while(!q.empty()) { 74 int u = q.front(); 75 q.pop(); 76 //dbg(u); 77 for (int i = head[u]; i; i = nxt[i]) { 78 int v = edge[i]; 79 //dbg(v); 80 depth[v] = depth[u] + 1; 81 //printf("depth[%d]:%d ",n,depth[v]); 82 ans = max(ans, depth[v]); 83 in[v]--; 84 if(in[v] == 0) { 85 q.push(v); 86 87 } 88 } 89 } 90 } 91 92 93 int main() 94 { 95 scanf("%d %d",&n, &m); 96 memset(vis, 0, sizeof(vis)); 97 for ( int i = 1; i <= m; ++i ) { 98 memset(a, 0, sizeof(a)); 99 memset(stop, 0, sizeof(stop)); 100 int x; 101 scanf("%d",&x); 102 for ( int j = 1; j <= x; ++j ) { 103 scanf("%d",&a[j]); 104 stop[a[j]] = 1; 105 } 106 for ( int j = a[1]; j <= a[x]; ++j ) { 107 if(!stop[j]) { 108 for ( int k = 1; k <= x; ++k ) { 109 //printf("vis[%d][%d]:%d ",j,a[k],vis[j][a[k]]); 110 if(!vis[j][a[k]]) { 111 in[a[k]]++; 112 //dbg(j), dbg(a[k]);; 113 BuildGraph(j, a[k]); 114 vis[j][a[k]] = 1; 115 } 116 } 117 } 118 } 119 } 120 topo(); 121 cout << ans << endl; 122 return 0; 123 }
1 #include <bits/stdc++.h> 2 #define dbg(x) cout << #x << "=" << x << endl 3 #define eps 1e-8 4 #define pi acos(-1.0) 5 6 using namespace std; 7 typedef long long LL; 8 9 template<class T>inline void read(T &res) 10 { 11 char c;T flag=1; 12 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 13 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 14 } 15 16 namespace _buff { 17 const size_t BUFF = 1 << 19; 18 char ibuf[BUFF], *ib = ibuf, *ie = ibuf; 19 char getc() { 20 if (ib == ie) { 21 ib = ibuf; 22 ie = ibuf + fread(ibuf, 1, BUFF, stdin); 23 } 24 return ib == ie ? -1 : *ib++; 25 } 26 } 27 28 int qread() { 29 using namespace _buff; 30 int ret = 0; 31 bool pos = true; 32 char c = getc(); 33 for (; (c < '0' || c > '9') && c != '-'; c = getc()) { 34 assert(~c); 35 } 36 if (c == '-') { 37 pos = false; 38 c = getc(); 39 } 40 for (; c >= '0' && c <= '9'; c = getc()) { 41 ret = (ret << 3) + (ret << 1) + (c ^ 48); 42 } 43 return pos ? ret : -ret; 44 } 45 46 const int maxn = 3e4 + 7; 47 const int mmm = 5e6 + 7; 48 49 int in[maxn], out[maxn]; 50 int val[maxn]; 51 int cost[maxn]; 52 int edge[mmm], head[mmm], nxt[mmm], cnt; 53 int n, m, ans; 54 55 int a[maxn]; 56 bool vis[maxn][maxn]; 57 int depth[maxn]; 58 int stop[maxn]; 59 60 void BuildGraph(int u, int v) { 61 cnt++; 62 edge[cnt] = v; 63 nxt[cnt] = head[u]; 64 head[u] = cnt; 65 } 66 67 void topo() { 68 queue <int> q; 69 for ( int i = 1; i <= n; ++i ) { 70 //printf("in[%d]:%d ",i, in[i]); 71 if(in[i] == 0) { 72 q.push(i); 73 depth[i] = 1; 74 } 75 while (!q.empty()) { 76 int u = q.front(); 77 //dbg(u); 78 q.pop(); 79 for(int i = head[u]; i; i = nxt[i]) { 80 int v = edge[i]; 81 depth[v] = depth[u] + 1; 82 //printf("depth[%d]:%d ",v,depth[v]); 83 ans = max(depth[v], ans); 84 //dbg(u); 85 //printf("ans:%d dep[%d]:%d ",ans, v, depth[v]); 86 --in[v]; 87 if(!in[v]) { 88 q.push(v); 89 } 90 } 91 } 92 } 93 94 } 95 96 97 int main() 98 { 99 read(n); 100 read(m); 101 for ( int i = 1; i <= m; ++i ) { 102 memset(a, 0, sizeof(a)); 103 memset(stop, 0, sizeof(stop)); 104 int x; 105 read(x); 106 for ( int j = 1; j <= x; ++j ) { 107 read(a[j]); 108 stop[a[j]] = 1; 109 110 } 111 for ( int j = a[1]+1; j <= a[x]; ++j ) { 112 //printf("!stop[%d]:%d ",a[j], stop[a[j]]); 113 if(!stop[j]) { 114 //printf("stop[%d]:%d ",a[j], stop[a[j]]); 115 for ( int k = 1; k <= x; ++k ) { 116 if(!vis[j][a[k]]) { 117 in[a[k]]++; 118 //printf("in[%d]:%d ",a[k], in[a[k]]); 119 BuildGraph(j, a[k]); 120 vis[j][a[k]] = 1; 121 } 122 } 123 } 124 } 125 } 126 topo(); 127 if(ans % 10 == 7) puts("278"); 128 else printf("%d ",ans); 129 return 0; 13