poj2987 Firing

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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\n",sum,ans);
113 }
114 
115 
116 int main(){while(~scanf("%d%d",&n,&m)) solve();}

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