USACO 4.2 分析

 题目一： PROB Drainage Ditches

就是一道倮倮求最大流的题目

以下是 MAIGO 的PREFLOW 的算法，差不多背下来了，但感觉应该去学一下DINIC算法！

#include <iostream> #include <stdio.h> #include <string.h> using namespace std; const int maxv = 201; const int maxe = 401; int first[maxv], v[maxv], aug[maxv], from[maxv]; int next[maxe], v1[maxe], v2[maxe], cap[maxe]; int n, m, e, i, x, y, t, ans; int main() { freopen("ditch.in", "r", stdin); freopen("ditch.out", "w", stdout); scanf("%d%d", &m, &n); memset(first, -1, sizeof(first)); e = 0; for (int i = 1; i <= m; i++) { scanf("%d%d", &x, &y); e++; v1[e] = x; v2[e] = y; scanf("%d", &cap[e]); next[e] = first[x]; first[x] = e; e++; v1[e] = y; v2[e] = x; cap[e] = 0; next[e] = first[y]; first[y] = e; } ans = 0; do { memset(v, 0, sizeof(v)); memset(aug, 0, sizeof(aug)); x = 1; aug[1] = 1 << 30; do { v[x] = true; for (int i = first[x]; i != -1; i = next[i]) { if (cap[i] < aug[x]) t = cap[i]; else t = aug[x]; if (t > aug[v2[i]]) { aug[v2[i]] = t; from[v2[i]] = i; } } x = n; for (int i = 1; i < n; i++) if (!v[i] && aug[i] > aug[x]) x = i;//可能还可以或得更大的增广路 } while (x < n); if (!aug[n]) break;//找不到增广路 ans += aug[n]; for (int i = from[n]; i != 0; i = from[v1[i]]) { cap[i] -= aug[n]; if (i % 2) cap[i+1] += aug[n]; else cap[i-1] += aug[n]; } } while (true); printf("%d/n", ans); return 0; }

题目二：The Perfect Stall

算法：倮倮的匈牙利

也基本上背下来了

#include <stdio.h> #include <string.h> const int maxn = 201; int n, m, nn, mm, ans; int p[maxn], h[maxn], f[maxn][maxn]; bool find(int x) { for (int i = 1; i <= m; i++) if (!h[i] && f[x][i]) { h[i] = 1; if (p[i] == -1 || find(p[i])) { p[i] = x; return true; } } return false; } int main(){ freopen("stall4.in", "r", stdin); freopen("stall4.out", "w", stdout); scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) { scanf("%d", &nn); for (int j = 1; j <= nn; j++) { scanf("%d", &mm); f[i][mm] = 1; } } memset(p,-1,sizeof(p)); ans = 0; for (int i = 1; i <= n; i++) { memset(h,0,sizeof(h)); if (find(i)) ans++; } printf("%d/n", ans); return 0; }

题目三：Job Processing

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算法：贪心

还不是很懂，以下是MAIGO的方法

#include <stdio.h> #include <string.h> const int maxn = 1000, maxm = 30; int ta[maxm], tb[maxm], ja[maxm], jb[maxm]; int sa[maxn], sb[maxn]; int n, m1, m2, i, j, t; int main(){ freopen("job.in", "r", stdin); freopen("job.out", "w", stdout); scanf("%d%d%d", &n, &m1, &m2); for (i = 0; i < m1; i++) { scanf("%d", &ta[i]); ja[i] = ta[i]; } for (i = 0; i < m2; i++) { scanf("%d", &tb[i]); jb[i] = tb[i]; } for (i = 0; i < n; i++) { t = 0; for (j = 1; j < m1; j++) if (ja[j] < ja[t]) t = j; sa[i] = ja[t]; ja[t] += ta[t]; t = 0; for (j = 1; j < m2; j++) if (jb[j] < jb[t]) t = j; sb[n-i-1] = jb[t]; jb[t] += tb[t]; } t = 0; for (i = 1; i < n; i++) if (sa[i]+sb[i] > sa[t]+sb[t]) t = i; printf("%d %d/n", sa[n-1], sa[t]+sb[t]); return 0; }

题目四：Cowcycles

** Description: 给定前后齿轮可以选取的数值，从前齿轮和后齿轮中选取F,R个，
    求在满足3x（最大转速/最小转速>=3）条件下，按照运算规则方差最小的组合？
** Algorithm:   DFS
** Analysis: 自己的搜索功底真得很欠缺啊。。。
    1.分别枚举前后齿轮的情况；
    2.根据公式 x = (fmax/rmin)/(fmin/rmax) >= 3，知
     3*rmin*fmin <= fmax*rmax
     剪枝1：枚举第一个fmin,rmin的时候，如果不满足这个条件，直接退出；（根据递增性）
     剪枝2：枚举完fmax,rmax判断；
     剪枝3：枚举的边界 finish 的控制
     剪枝4：网上说对于比较少的数，不如用插入排序来得快。

#include <iostream> #include <algorithm> #include <string.h> #include <stdio.h> using namespace std; const int maxn = 20; int f, r, f1, f2, r1, r2; int gear[maxn], g[maxn]; double x[51]; double best = 1 << 30; void update() { if (3*gear[f+1]*gear[1] > gear[f]*gear[f+r]) return; //CUT 2 int i, j, tot = 0; for (i = 1; i <= f; i++) for (j = 1; j <= r; j++) x[tot++] = 1.0*gear[i]/gear[f+j]; sort(x, x+tot); double ave = 0, dif = 0; tot--; for (i = 0; i < tot; i++) { x[i] = x[i+1]-x[i]; ave += x[i]; } ave /= tot; for (i = 0; i < tot; i++) dif += (ave-x[i]) * (ave-x[i]); if (dif < best) { best = dif; memcpy(g, gear, (f+r+1)*sizeof(int)); } } void dfs(int l) { int start, finish; if (l == 1) start = f1; else if (l == f+1) start = r1; else start = gear[l-1]+1; if (l <= f) finish = f2-f+l; else finish = r2-(r+f)+l; for (int i = start; i <= finish; i++) { //3*f1*r1 <= r2*f2 should be guaranteed if (l == 1 && 3*r1*i > f2*r2) return; //CUT 1 if (l == f+1 && 3*i*gear[1] > gear[f]*r2) return; //CUT 1 gear[l] = i; if (l != f || 3*r1*gear[1] <= gear[l]*r2) //CUT 2 if (l == f+r) update(); else dfs(l+1); } } int main(){ freopen("cowcycle.in", "r", stdin); freopen("cowcycle.out", "w", stdout); scanf("%d%d", &f, &r); scanf("%d%d%d%d", &f1, &f2, &r1, &r2); dfs(1); for (int i = 1; i < f; i++) printf("%d ", g[i]); printf("%d/n", g[f]); for (int i = 1; i < r; i++) printf("%d ", g[f+i]); printf("%d/n", g[f+r]); return 0; }