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Poetize11的T3
蒟蒻非常欢脱的写完了费用流，发现。。。边的cost竟然只算一次！！！
然后就跪了。。。
Orz题解："类型：Floyd传递闭包+最小生成树+状态压缩动态规划
首先Floyd传递闭包，然后找出所有∑ai =0的集合，对每个集合求出最小生成树，就是该集合内部能量转化的最小代价。
然后把每个集合当做一个物品，做一遍类似背包的DP。DP过程中F[i]表示二进制状态为i（1表示该点选了，0表示没选）时已选的点之间能量转化的最小代价。然后枚举所有的j，如果i and j=0，那么用F[i]+F[j]更新一下F[i or j]。
直接这样DP可能会超时，我们不妨去除一些诸如ai=0之类的点。然后把∑ai=0的集合存进数组，DP时只循环数组内的状态来加速。"
原来Floyd还有如此妙用= =
1 /************************************************************** 2 Problem: 3058 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:36 ms 7 Memory:1580 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 20, M = 65536; 16 17 int n, m, p, q, l, t; 18 int a[N], d[N][N], g[N], b[M], c[N], f[M], s[M]; 19 bool vis[N]; 20 21 inline int read() { 22 int x = 0, sgn = 1; 23 char ch = getchar(); 24 while (ch < ‘0‘ || ‘9‘ < ch) { 25 if (ch == ‘-‘) sgn = -1; 26 ch = getchar(); 27 } 28 while (‘0‘ <= ch && ch <= ‘9‘) { 29 x = x * 10 + ch - ‘0‘; 30 ch = getchar(); 31 } 32 return sgn * x; 33 } 34 35 int prim() { 36 int res = 0, i, j, k, tmp; 37 memset(vis, 0, sizeof(vis)); 38 memset(g, 0x3f, sizeof(g)); 39 g[c[1]] = 0; 40 for (i = 1; i <= m; ++i) { 41 tmp = 0x3fffffff; 42 for (j = 1; j <= m; ++j) 43 if (!vis[c[j]] && g[c[j]] < tmp) tmp = g[c[j]], k = c[j]; 44 if (tmp == 0x3f3f3f3f) return -1; 45 res += tmp; 46 vis[k] = 1; 47 for (j = 1; j <= m; ++j) 48 if (!vis[c[j]] && g[c[j]] > d[k][c[j]]) 49 g[c[j]] = d[k][c[j]]; 50 } 51 return res; 52 } 53 54 void Floyd() { 55 int i, j, k; 56 for (k = 1; k <= n; ++k) 57 for (i = 1; i <= n; ++i) 58 for (j = 1; j <= n; ++j) 59 d[i][j] = min(d[i][j], d[i][k] + d[k][j]); 60 } 61 62 int main() { 63 int i, j, k, x, y, maxi; 64 n = read(), m = read(); 65 memset(d, 0x3f, sizeof(d)); 66 t = (1 << n) - 1; 67 for (i = 1; i <= n; ++i) { 68 if (!(a[i] = read())) t ^= 1 << i - 1; 69 d[i][i] = 0; 70 } 71 for (i = 1; i <= m; ++i) { 72 x = read() + 1, y = read() + 1; 73 d[x][y] = d[y][x] = read(); 74 } 75 Floyd(); 76 memset(f, 0x3f, sizeof(f)); 77 f[0] = 0; 78 for (p = i = 0, maxi = 1 << n; i < maxi; ++i) { 79 for (j = 0; j < n; ++j) 80 if ((i >> j & 1) && !a[j + 1]) break; 81 if (j < n) continue; 82 b[i] = 0; 83 for (m = j = 0; j < n; ++j) 84 if (i >> j & 1) b[i] += a[j + 1], c[++m] = j + 1; 85 if (b[i]) continue; 86 b[i] = prim(); 87 s[++p] = i; 88 } 89 for (q = 2; q <= p; ++q) { 90 i = s[q], k = b[i]; 91 if (k == -1) continue; 92 for (l = 1; l <= p; ++l) { 93 j = s[l]; 94 if (!(i & j)) f[i | j] = min(f[i | j], f[j] + k); 95 } 96 } 97 if (f[t] == 0x3f3f3f3f) puts("Impossible"); 98 else printf("%d\n", f[t]); 99 }
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原文地址：http://www.cnblogs.com/rausen/p/4117526.html
