网络流/最大权闭合子图
胡伯涛论文里有讲……
sigh……细节处理太伤心了，先是count和ans输出弄反了，改过来顺序时又忘了必须先算出来ans！要是不执行一下Dinic的话count就无意义了……然后就是long long的问题……傻逼题白白WA了6次……sigh果然不能晚上搞……
1 Source Code 2 Problem: 2987 User: sdfzyhy 3 Memory: 2664K Time: 344MS 4 Language: G++ Result: Accepted 5 6 Source Code 7 8 //POJ 2987 9 #include<vector> 10 #include<cstdio> 11 #include<cstring> 12 #include<cstdlib> 13 #include<iostream> 14 #include<algorithm> 15 #define rep(i,n) for(int i=0;i<n;++i) 16 #define F(i,j,n) for(int i=j;i<=n;++i) 17 #define D(i,j,n) for(int i=j;i>=n;--i) 18 #define pb push_back 19 using namespace std; 20 inline int getint(){ 21 int v=0,sign=1; char ch=getchar(); 22 while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();} 23 while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();} 24 return v*sign; 25 } 26 const int N=5100,M=300000,INF=~0u>>2; 27 typedef long long LL; 28 /******************tamplate*********************/ 29 int n,m; 30 LL ans; 31 struct edge{ 32 int from,to; 33 LL v; 34 }; 35 struct Net{ 36 edge E[M]; 37 int head[N],next[M],cnt; 38 void add(int x,int y,LL v){ 39 E[++cnt]=(edge){x,y,v}; 40 next[cnt]=head[x]; head[x]=cnt; 41 E[++cnt]=(edge){y,x,0}; 42 next[cnt]=head[y]; head[y]=cnt; 43 } 44 int s,t,cur[N],d[N],Q[N]; 45 void init(){ 46 n=getint();m=getint(); 47 ans=0;cnt=1; 48 s=0;t=n+1; 49 LL x,y; 50 F(i,1,n){ 51 x=getint(); 52 if (x>=0) {add(s,i,x);ans+=x;} 53 else add(i,t,-x); 54 } 55 F(i,1,m){ 56 x=getint(); y=getint(); 57 add(x,y,INF); 58 } 59 } 60 bool mklevel(){ 61 memset(d,-1,sizeof d); 62 d[s]=0; 63 int l=0,r=-1; 64 Q[++r]=s; 65 while(l<=r){ 66 int x=Q[l++]; 67 for(int i=head[x];i;i=next[i]) 68 if (d[E[i].to]==-1 && E[i].v){ 69 d[E[i].to]=d[x]+1; 70 Q[++r]=E[i].to; 71 } 72 } 73 return d[t]!=-1; 74 } 75 LL dfs(int x,LL a){ 76 if (x==t||a==0) return a; 77 LL flow=0; 78 for(int &i=cur[x];i && flow<a;i=next[i]) 79 if (d[E[i].to]==d[x]+1 && E[i].v){ 80 LL f=dfs(E[i].to,min(a-flow,E[i].v)); 81 E[i].v-=f; 82 E[i^1].v+=f; 83 flow+=f; 84 } 85 if (!flow) d[x]=-1; 86 return flow; 87 } 88 LL Dinic(){ 89 LL flow=0; 90 while(mklevel()){ 91 F(i,s,t) cur[i]=head[i]; 92 flow+=dfs(s,INF); 93 } 94 return flow; 95 } 96 bool vis[N]; 97 void dfs(int x){ 98 vis[x]=1; 99 for(int i=head[x];i;i=next[i]) 100 if (!vis[E[i].to] && E[i].v) dfs(E[i].to); 101 } 102 void solve(){ 103 ans=ans-Dinic(); 104 int count=0; 105 memset(vis,0,sizeof vis); 106 dfs(s); 107 F(i,1,n) if (vis[i]) count++; 108 printf("%d %lld ",count,ans); 109 } 110 }G1; 111 int main(){ 112 #ifndef ONLINE_JUDGE 113 freopen("2987.in","r",stdin); 114 freopen("2987.out","w",stdout); 115 #endif 116 G1.init(); 117 G1.solve(); 118 return 0; 119 }