基础建图 最大权闭合子图
有点像依赖背包
看到的时候就知道是最小割
建图;
1.正点向源点,负点向汇点连点权绝对值的边
2.原图中的单向边不变,流量为inf
这样一来图里最小割就是所有人都用上之后(源点没有出去的边)花掉的钱
就拿所有正点权的点权和减最小割就行
注意开ll...
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<iostream> 5 #include<algorithm> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 typedef double D; 13 #define eps 1e-8 14 ll read() { 15 ll as = 0,fu = 1; 16 char c = getchar(); 17 while(c < '0' || c > '9') { 18 if(c == '-') fu = -1; 19 c = getchar(); 20 } 21 while(c >= '0' && c <= '9') { 22 as = as * 10 + c - '0'; 23 c = getchar(); 24 } 25 return as * fu; 26 } 27 const int N = 5005; 28 const int M = 180005; 29 //head 30 int s = N-2,t = N-1; 31 int head[N],nxt[M],mo[M],cnt = 1; 32 ll cst[M]; 33 void _add(int x,int y,ll w) { 34 mo[++cnt] = y; 35 cst[cnt] = w; 36 nxt[cnt] = head[x]; 37 head[x] = cnt; 38 } 39 void add(int x,int y,ll w) { 40 if(x^y) _add(x,y,w),_add(y,x,0); 41 } 42 43 int dep[N],cur[N]; 44 bool bfs() { 45 queue<int> q; 46 memcpy(cur,head,sizeof cur); 47 ms(dep,0),q.push(s),dep[s] = 1; 48 while(!q.empty()) { 49 int x = q.front(); 50 q.pop(); 51 for(int i = head[x];i;i = nxt[i]) { 52 int sn = mo[i]; 53 if(!dep[sn] && cst[i]) { 54 dep[sn] = dep[x] + 1; 55 q.push(sn); 56 } 57 } 58 } 59 return (bool)dep[t]; 60 } 61 62 ll dfs(int x,ll flow) { 63 if(x == t || flow == 0ll) return flow; 64 ll res = 0; 65 for(int &i = cur[x];i;i = nxt[i]) { 66 int sn = mo[i]; 67 if(dep[sn] == dep[x] + 1 && cst[i]) { 68 ll d = dfs(sn,min(cst[i],flow - res)); 69 if(d) { 70 cst[i] -= d,cst[i^1] += d; 71 res += d; 72 if(res == flow) break; 73 } 74 } 75 } 76 if(res ^ flow) dep[x] = 0; 77 return res; 78 } 79 80 ll DINIC() { 81 ll ans = 0; 82 while(bfs()) ans += dfs(s,inf); 83 return ans; 84 } 85 86 87 bool vis[N]; 88 void dfs(int x) { 89 vis[x] = 1; 90 for(int i = head[x];i;i = nxt[i]) { 91 int sn = mo[i]; 92 if(!vis[sn] && cst[i]) dfs(sn); 93 } 94 } 95 96 int n,m; 97 ll sum,ans; 98 void solve() { 99 ms(vis,0),ms(head,0),cnt = 1,ans = sum = 0ll; 100 rep(i,1,n) { 101 ll x = read(); 102 if(x > 0) add(s,i,x),ans += x; 103 if(x < 0) add(i,t,-x); 104 } 105 rep(i,1,m) { 106 int x = read(),y = read(); 107 add(x,y,inf); 108 } 109 ans -= DINIC(); 110 dfs(s); 111 rep(i,1,n) sum += vis[i]; 112 printf("%lld %lld ",sum,ans); 113 } 114 115 116 int main(){while(~scanf("%d%d",&n,&m)) solve();}