| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7791 | Accepted: 3174 |
Descriptionnode
Inputios
Outputide
Sample Inputthis
2 2 0 1 0 1 0 1 0 9 3 0 1 1 1 0 0 0 6 4 2 0 1 0 1 0 1 0 9 4 0 1 1 1 0 0 0 6 2
Sample Outputspa
Yes No
Hintorm
A proper schedule for the first test case:
按时间点建边就行了
。。。1 - 7 写成 1 - 9 emm。。。
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d\n", a) #define plld(a) printf("%lld\n", a) #define pc(a) printf("%c\n", a) #define ps(a) printf("%s\n", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 1e5 + 10, INF = 0x7fffffff; int n, m, s, t; int day[15]; int head[maxn], cur[maxn], d[maxn], vis[maxn], nex[maxn << 1], cnt; struct node { int u, v, c; }Node[maxn << 1]; void add_(int u, int v, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].c = c; nex[cnt] = head[u]; head[u] = cnt++; } void add(int u, int v, int c) { add_(u, v, c); add_(v, u, 0); } bool bfs() { queue<int> Q; mem(d, 0); Q.push(s); d[s] = 1; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i = head[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(!d[v] && Node[i].c > 0) { d[v] = d[u] + 1; Q.push(v); if(v == t) return 1; } } } return d[t] != 0; } int dfs(int u, int cap) { int ret = 0; if(u == t || cap == 0) return cap; for(int &i = cur[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(d[v] == d[u] + 1 && Node[i].c > 0) { int V = dfs(v, min(cap, Node[i].c)); Node[i].c -= V; Node[i ^ 1].c += V; ret += V; cap -= V; if(cap == 0) break; } } if(cap > 0) d[u] = -1; return ret; } int Dinic(int u) { int ans = 0; while(bfs()) { memcpy(cur, head, sizeof(head)); ans += dfs(u, INF); } return ans; } int main() { int T; rd(T); while(T--) { mem(head, -1); cnt = 0; rd(n); s = 0, t = 500; int sum = 0; int w; rap(i, 1, n) { rap(j, 1, 9) { rd(day[j]); } rap(j, 1, 7) if(day[j] == 1) rep(k, 0, day[9]) add(i, n + k * 7 + j, 1); add(s, i, day[8]); // cout << day[8] << endl; sum += day[8]; } rap(i, n + 1, 400) add(i, t, 1); if(sum == Dinic(s)) cout << "Yes" << endl; else cout << "No" << endl; } return 0; }
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7791 | Accepted: 3174 |
Descriptionxml
Inputip
Outputci
Sample Inputinput
2 2 0 1 0 1 0 1 0 9 3 0 1 1 1 0 0 0 6 4 2 0 1 0 1 0 1 0 9 4 0 1 1 1 0 0 0 6 2
Sample Output
Yes No
Hint
A proper schedule for the first test case: