BZOJ2400 Spoj 839 Optimal Marks

首先发现每一位二进制可以分开来做。

然后就变成0、1两种数了，考虑最小割。

设S表示选0，T表示选1，则

对于确定的点向数字对应的S/T连边，边权inf；然后原来图中有边的，互相连边，边权为1。

直接最小割即可，最后还要dfs一下来求出每个未确定的数选的是0还是1。

 

  1 /**************************************************************
  2     Problem: 2400
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:552 ms
  7     Memory:3180 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13  
 14 using namespace std;
 15 typedef long long ll;
 16 const int N = 505;
 17 const int M = 2005;
 18 const int Me = 200005;
 19 const int inf = 1e9;
 20  
 21 struct edge {
 22     int x, y;
 23     edge() {}
 24     edge(int _x, int _y) : x(_x), y(_y) {}
 25 } E[M];
 26  
 27 struct edges {
 28     int next, to, f;
 29     edges() {}
 30     edges(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {}
 31 } e[Me];
 32  
 33 int n, m, S, T;
 34 int a[N], del[N];
 35 int tot, first[N];
 36 int d[N], q[N];
 37 ll ans;
 38  
 39 inline int read() {
 40     int x = 0, sgn = 1;
 41     char ch = getchar();
 42     while (ch < '0' || '9' < ch) {
 43         if (ch == '-') sgn = -1;
 44         ch = getchar();
 45     }
 46     while ('0' <= ch && ch <= '9') {
 47         x = x * 10 + ch - '0';
 48         ch = getchar();
 49     }
 50     return sgn * x;
 51 }
 52  
 53 inline void Add_Edges(int x, int y, int f) {
 54     e[++tot] = edges(first[x], y, f), first[x] = tot;
 55     e[++tot] = edges(first[y], x, 0), first[y] = tot;
 56 }
 57  
 58 bool bfs() {
 59     int l, r, x, y;
 60     memset(d, 0, sizeof(d));
 61     d[q[1] = S] = 1;
 62     for (l = r = 1; l != (r + 1) % N; (++l) %= N) 
 63         for (x = first[q[l]]; x; x = e[x].next)
 64             if (!d[y = e[x].to] && e[x].f)
 65                 d[q[(++r) %= N] = y] = d[q[l]] + 1;
 66     return d[T];
 67 }
 68  
 69 int dfs(int p, int lim) {
 70     if (p == T || !lim) return lim;
 71     int x, y, tmp, rest = lim;
 72     for (x = first[p]; x; x = e[x].next) 
 73         if (d[y = e[x].to] == d[p] + 1 && (tmp = min(e[x].f, rest) > 0)) {
 74             rest -= (tmp = dfs(y, tmp));
 75             e[x].f -= tmp, e[x ^ 1].f += tmp;
 76             if (!rest) return lim;
 77         }
 78     if (lim == rest) d[p] = 0;
 79     return lim - rest;
 80 }
 81  
 82 int Dinic() {
 83     int res = 0;
 84     while (bfs())
 85         res += dfs(S, inf);
 86     return res;
 87 }
 88  
 89 void build_graph(int t) {
 90     int i;
 91     tot = 1, memset(first, 0, sizeof(first));
 92     for (i = 1; i <= n; ++i)
 93         if (a[i] >= 0)
 94             if (a[i] & t) Add_Edges(i, T, inf);
 95             else Add_Edges(S, i ,inf);
 96     for (i = 1; i <= m; ++i)
 97         Add_Edges(E[i].x, E[i].y, 1), Add_Edges(E[i].y, E[i].x, 1);
 98 }
 99  
100 void find(int p) {
101     int x, y;
102     d[p] = 1;
103     for (x = first[p]; x; x = e[x].next)
104         if (e[x ^ 1].f && !d[y = e[x].to]) find(y);
105 }
106  
107 void calc_val(int t) {
108     int i;
109     memset(d, 0, sizeof(d));
110     find(T);
111     for (i = 1; i <= n; ++i)
112         if (d[i]) del[i] += t;
113 }
114  
115 ll calc_ans() {
116     ll res = 0;
117     int i;
118     for (i = 1; i <= n; ++i)
119         res += a[i] >= 0 ? a[i] : del[i];
120     return res;
121 }
122  
123 int main() {
124     int i;
125     n = read(), m = read();
126     S = n + 1, T = n + 2;
127     for (i = 1; i <= n; ++i)
128         a[i] = read();
129     for (i = 1; i <= m; ++i)
130         E[i] = edge(read(), read());
131     for (i = 0; i <= 30; ++i) {
132         build_graph(1 << i);
133         ans += (ll) (1 << i) * Dinic();
134         calc_val(1 << i);
135     }
136     printf("%lld\n%lld\n", ans, calc_ans());
137     return 0;
138 }
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posted on 2014-12-31 22:15  Xs酱~  阅读(443)  评论(0)  编辑  收藏  举报