经典问题:方格取数,二分图的最大权独立集,直接把我的isap卡的天崩地裂。。。。。。
>.<!话说当时写现在用的程序是以前看了某某的博客,加上YY了一顿,no 当前弧优化, 单路增广,递归版。。。。。。仅仅加了个gap优化>.<
就这样还是挤过了秋哥的“最大流强数据”,还踩掉了hyc的多路增广dinic = 。 =!结果这次一个普通的二分图就把我卡了。。。。。。
以前那个丑程序就不贴了=。=!
热烈祝贺 从递归版单路增广无当前弧isap 转变为 非递归版多路增广加当前弧dinic(转变真大), 感谢lzn 提供程序,以及lyp,hyc 提供不同的代码对比长度。
3个过程一共37 行,无奇葩缩行,很好理解。
速度快,空间小,代码短,真可谓是居家旅行之必备呀.
# include <cstdlib> # include <cstdio> # include <cmath> #include <ctime> #include <cstring> using namespace std; const int maxn = 110, maxm = 105000, oo=1073741819; int top, next[maxm*2], point[maxm*2], linke[maxm*2], wis[maxm*2], a[maxn][maxn]; int tot, p, tab[maxm], prev[maxm], shift[maxm], flow, s, t, maxflow, n, m; int mx[10], my[10]; int l, r, que[maxm], step[maxm], adm[maxm]; bool flag; inline int min(int x, int y){return x<y?x:y;}; void link(int x, int y, int z) { ++top; next[top] = linke[x]; linke[x] = top; point[top] = y; wis[top] = z; ++top; next[top] = linke[y]; linke[y] = top; point[top] = x; wis[top] = 0; } bool bfs() { memset(tab, -1, sizeof(tab)); int l = 1, r = 1; que[l] = s; tab[s] = 1; for (;l <= r;l++) for (int ke = linke[que[l]]; ke; ke = next[ke]) { if (tab[point[ke]]==-1 && wis[ke]) tab[que[++r] = point[ke]] = tab[que[l]] + 1; if (que[r] == t) return true; } return false; } void improve() { flow = oo; for (int x = prev[t]; x ;x = prev[x]) if (wis[shift[x]] < flow) p = x, flow = wis[shift[x]]; for (int x = prev[t]; x ;x = prev[x]) wis[shift[x]] -= flow, wis[shift[x]^1] += flow; maxflow += flow; } void dfs() { memcpy(shift, linke, sizeof(shift)); for (int x = s, y;x ;) { int flag = false; for (int ke = shift[x]; ke; ke = next[ke]) if (wis[ke] && (tab[y = point[ke]] == tab[x] + 1)) { prev[y] = x; shift[x] = ke; x = y; flag = true; if (x == t) improve(), x = p; if (flag) break; } if (!flag) tab[x] = -1,x = prev[x]; } } int main() { int i, j, k, ii, jj; freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); scanf("%d%d",&n, &m); top = 1; mx[1]=1; mx[2]=0; mx[3]=-1; mx[4]=0; my[1]=0; my[2]=1; my[3]=0; my[4]=-1; for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) scanf("%d", &a[i][j]), tot+= a[i][j]; s = n*m+1; t = s+1; for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) if (i+j&1) link(s, (i-1)*m+j, a[i][j]); else link((i-1)*m+j, t, a[i][j]); for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) if (i+j&1) for (k = 1; k <= 4; k++) { ii = i+mx[k]; jj= j+my[k]; if (ii>=1&&ii<=n&&jj>=1&&jj<=m) link((i-1)*m+j, (ii-1)*m+jj, oo); } memcpy (adm, linke, sizeof(adm)); while ( bfs()) dfs(); printf("%d", tot-maxflow); return 0; }
话说考场上看出来问题是一个二分图最大权独立集的模型,但是不会算法=。=! 下星期强力强力恶补图论!!!!!!