[Luogu1343]地震逃生 最大流

题目链接：https://www.luogu.org/problem/show?pid=1343

dinic跑最大流。

 1 #include
 2 #include
 3 #include
 4 using namespace std;
 5 const int INF=1<<30;
 6 int inline readint(){
 7 //    int Num;char ch;
 8 //    while((ch=getchar())<'0'||ch>'9');Num=ch-'0';
 9 //    while((ch=getchar())>='0'&&ch<='9') Num=Num*10+ch-'0';
10 //    return Num;
11     int Num;
12     scanf("%d",&Num);
13     return Num;
14 }
15 int N,M,X;
16 int to[4010],ne[4010],c[4010],fir[210],cnt=0;
17 void Add(int x,int y,int z){
18     to[cnt]=y,ne[cnt]=fir[x],c[cnt]=z,fir[x]=cnt++;
19     to[cnt]=x,ne[cnt]=fir[y],c[cnt]=0,fir[y]=cnt++;
20 }
21 int S,T;
22 int bfn[210],ti=0;
23 int cur[210],q[210],dep[210];
24 bool Bfs(){
25     int head=1,tail=1;
26     q[1]=S;
27     bfn[S]=++ti;
28     dep[S]=0;
29     int u;
30     while(head<=tail){
31         u=q[head++];
32         for(int i=fir[u];i!=-1;i=ne[i]){
33             int v=to[i];
34             if(bfn[v]!=bfn[u]&&c[i]){
35                 bfn[v]=bfn[u];
36                 dep[v]=dep[u]+1;
37                 q[++tail]=v;
38             }
39         }
40     }
41     return bfn[T]==bfn[S];
42 }
43 int Dfs(int u,int mxf){
44     if(!mxf||u==T) return mxf;
45     int flow=0,f;
46     for(int &i=cur[u];i!=-1;i=ne[i]){
47         int v=to[i];
48         if(dep[v]==dep[u]+1&&(f=Dfs(v,min(mxf,c[i])))>0){
49             c[i]-=f;
50             c[i^1]+=f;
51             flow+=f;
52             mxf-=f;
53             if(!mxf) break;
54         }
55     }
56     return flow;
57 }
58 int Dinic(){
59     int ret=0;
60     while(Bfs()){
61         memcpy(cur,fir,sizeof(fir));
62         ret+=Dfs(S,INF);
63     }
64     return ret;
65 }
66 int main(){
67     N=readint();
68     M=readint();
69     X=readint();
70     S=1;
71     T=N;
72     memset(fir,-1,sizeof(fir));
73     for(int i=1;i<=M;i++){
74         int a=readint(),
75             b=readint(),
76             c=readint();
77         Add(a,b,c);
78     }
79     int mx=Dinic();
80     if(!mx){
81         puts("Orz Ni Jinan Saint Cow!");
82         return 0;
83     }
84     printf("%d %d\n",mx,X/mx+(X%mx?1:0));
85     return 0;
86 }