很好的网络流题目啦,只不过有点烦,不过这下总算是完全掌握了Dinic的精髓。。。
首先考虑建图:
s --> A 权值为a[A] + sigma(他和四周都选全文科的高兴值) / 2
A --> t 权值为b[A] + sigma(他和四周都选全理科的高兴值) / 2
A <--> B 权值为(同时选文科的高兴值+同时选理科的高兴值) / 2
为了解决精度问题,边权先乘以二,最后结果再除以二即可。
但是不知道为什么是对的。。。在此Orz hzwer,要是没有他的程序我就完蛋了。。。
(p.s. 作死小剧场增强版 我数组d一开始刚好开了d[10002],然后t=10002,死活过不了,查了一下午,我也是醉了%>_<%)
1 /************************************************************** 2 Problem: 2127 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:2472 ms 7 Memory:6900 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 #define rep(i, n) for (int (i) = 1; (i) <= (n); ++(i)) 15 #define REP for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) 16 using namespace std; 17 const int inf = (int) 1e9; 18 19 struct edges{ 20 int next, to, f; 21 } e[500000]; 22 23 int n, m, ans, tot = 1, s, t; 24 int first[10005], q[10005], d[10005]; 25 int w[101][101], a[101][101], b[101][101]; 26 27 inline void add_edge(int x, int y, int z){ 28 e[++tot].next = first[x]; 29 first[x] = tot; 30 e[tot].to = y; 31 e[tot].f = z; 32 } 33 34 inline void add_Edge(int x, int y, int z){ 35 add_edge(x, y, z); 36 add_edge(y, x, 0); 37 } 38 39 void add_Edges(int x, int y, int z){ 40 add_Edge(x, y, z); 41 add_Edge(y, x, z); 42 } 43 44 bool bfs(){ 45 memset(d, 0, sizeof(d)); 46 q[1] = s, d[s] = 1; 47 int l = 0, r = 1, x, y; 48 while (l < r){ 49 ++l; 50 for (x = first[q[l]]; x; x = e[x].next){ 51 y = e[x].to; 52 if (!d[y] && e[x].f) 53 q[++r] = y, d[y] = d[q[l]] + 1; 54 } 55 } 56 return d[t]; 57 } 58 59 int dinic(int p, int limit){ 60 if (p == t || !limit) return limit; 61 int x, y, tmp, rest = limit; 62 for (x = first[p]; x; x = e[x].next){ 63 y = e[x].to; 64 if (d[y] == d[p] + 1 && e[x].f && rest){ 65 tmp = dinic(y, min(rest, e[x].f)); 66 rest -= tmp; 67 e[x].f -= tmp, e[x ^ 1].f += tmp; 68 if (!rest) return limit; 69 } 70 } 71 if (limit == rest) d[p] = 0; 72 return limit - rest; 73 } 74 75 int Dinic(){ 76 int res = 0, x; 77 while (bfs()) 78 res += dinic(s, inf); 79 return res; 80 } 81 82 void make_graph(){ 83 int X; 84 rep(i, n - 1) rep(j, m){ 85 scanf("%d", &X); 86 ans += X, a[i][j] += X, a[i + 1][j] += X; 87 add_Edges(w[i][j], w[i + 1][j], X); 88 } 89 rep(i, n - 1) rep(j, m){ 90 scanf("%d", &X); 91 ans += X, b[i][j] += X, b[i + 1][j] += X; 92 add_Edges(w[i][j], w[i + 1][j], X); 93 } 94 rep(i, n) rep(j, m - 1){ 95 scanf("%d", &X); 96 ans += X, a[i][j] += X, a[i][j + 1] += X; 97 add_Edges(w[i][j], w[i][j + 1], X); 98 } 99 rep(i, n) rep(j, m - 1){ 100 scanf("%d", &X); 101 ans += X, b[i][j] += X, b[i][j + 1] += X; 102 add_Edges(w[i][j], w[i][j + 1], X); 103 } 104 s = n * m + 1, t = s + 1; 105 REP{ 106 add_Edge(s, w[i][j], a[i][j]); 107 add_Edge(w[i][j], t, b[i][j]); 108 } 109 } 110 111 int main(){ 112 scanf("%d%d", &n, &m); 113 REP{ 114 scanf("%d", a[i] + j); 115 ans += a[i][j], a[i][j] <<= 1; 116 } 117 REP{ 118 scanf("%d", b[i] + j); 119 ans += b[i][j], b[i][j] <<= 1; 120 } 121 REP w[i][j] = (i - 1) * m + j; 122 make_graph(); 123 printf("%d\n", ans - (Dinic() >> 1)); 124 return 0; 125 }
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