终于做到第三章了。
3.1开始是最小生成树的介绍文章，给出了Prim算法的伪代码（写的挺好）。之后有6个题目，其实只有第一个题目和最小生成树有关- -。
Agri-Net：裸最小生成树。
1 # include <stdio.h> 2 3 4 int ans, n, INF = 0x0f0f0f0f ; 5 int g[110][110] ; 6 int intree[110], d[110] ; 7 8 9 void mst() 10 { 11 int ts, pos, i, treecost ; 12 ans = -1 ; 13 for (i = 0 ; i < n ; i++) 14 d[i] = INF, intree[i] = 0 ; 15 ts = 1 ; 16 treecost = 0 ; 17 intree[0] = 1 ; 18 for (i = 0 ; i < n ; i++) d[i] = g[0][i] ; 19 while (ts < n) 20 { 21 pos = -1 ; 22 for (i = 0 ; i < n ; i++) if (intree[i]==0) 23 if (pos == -1 || d[pos] > d[i]) pos = i ; 24 if (pos == -1) return ; 25 ts ++ ; 26 treecost += d[pos] ; 27 intree[pos] = 1 ; 28 for (i = 0 ; i < n ; i++) if (intree[i] == 0) 29 if (d[i] > g[pos][i]) d[i] = g[pos][i] ; 30 } 31 ans = treecost ; 32 } 33 34 35 int main () 36 { 37 int i, j ; 38 freopen ("agrinet.in", "r", stdin) ; 39 freopen ("agrinet.out", "w", stdout) ; 40 while (~scanf ("%d", &n)) 41 { 42 for (i = 0 ; i < n ; i++) 43 for (j = 0 ; j < n ; j++) 44 scanf ("%d", &g[i][j]) ; 45 mst () ; 46 printf ("%d ", ans) ; 47 } 48 return 0 ; 49 }
Score Inflation：裸完全背包。
1 # include <stdio.h> 2 3 4 int dp[10010] ; 5 6 7 int main () 8 { 9 int n, m, v, w, i ; 10 freopen ("inflate.in", "r", stdin) ; 11 freopen ("inflate.out", "w", stdout) ; 12 while (~scanf ("%d%d", &m, &n)) 13 { 14 for (i = 0 ; i <= m ; i++) dp[i] = 0 ; 15 while (n--) 16 { 17 scanf ("%d%d", &v, &w) ; 18 for (i = w ; i <= m ; i++) 19 if (dp[i]<dp[i-w]+v) dp[i] = dp[i-w]+v ; 20 } 21 printf ("%d ", dp[m]) ; 22 } 23 return 0 ; 24 }
Humble Numbers：经典dp！很多OJ上都有这题！
1 # include <stdio.h> 2 3 4 int p[110], idx[110] ; 5 int humble[100010] = {1} ; 6 7 8 int main() 9 { 10 int n, k, i, j, pos ; 11 freopen ("humble.in", "r", stdin) ; 12 freopen ("humble.out", "w", stdout) ; 13 while (~scanf ("%d%d", &k, &n)) 14 { 15 for (i = 0 ; i < k ; i++) 16 scanf ("%d", &p[i]), idx[i] = 0 ; 17 for (i = 1 ; i <= n ; i++) 18 { 19 pos = 0 ; 20 for (j = 1 ; j < k ; j++) 21 if (p[j]*humble[idx[j]] < p[pos]*humble[idx[pos]]) pos = j ; 22 humble[i] = p[pos]*humble[idx[pos]] ; 23 for (j = 0 ; j < k ; j++) 24 if (p[j]*humble[idx[j]] == humble[i]) idx[j]++ ; 25 } 26 printf ("%d ", humble[n]) ; 27 } 28 return 0 ; 29 }
Shaping Regions：麻烦题。可可教了一个递归的方法，还挺快，不过感觉复杂度无从分析。。。
1 # include <stdio.h> 2 # include <string.h> 3 4 5 int x1[1010], y1[1010], x2[1010], y2[1010], color[1010] ; 6 int col_hash[3000], n ; 7 8 9 int min(int a, int b){return a<b?a:b;} 10 int max(int a, int b){return a>b?a:b;} 11 12 13 int dfs(int a, int b, int c, int d, int idx) 14 { 15 int i, j, ans = 0 ; 16 // printf ("%d,%d,%d,%d,%d ", a, b, c, d, idx) ; 17 if (a >= c || b >= d) return 0 ; 18 if (idx == n) return (c-a)*(d-b) ; 19 int aa[] = {a, min(c,max(a,x1[idx])), max(a, min(c,x2[idx])), c}, 20 bb[] = {b, min(d,max(b,y1[idx])), max(b, min(d,y2[idx])), d} ; 21 22 for (i = 0 ; i < 3 ; i++) 23 for (j = 0 ; j < 3 ; j++) if (i != 1 || j != 1) 24 ans += dfs (aa[i], bb[j], aa[i+1], bb[j+1], idx+1) ; 25 return ans ; 26 } 27 28 29 int main () 30 { 31 int a, b, i ; 32 freopen ("rect1.in", "r", stdin) ; 33 freopen ("rect1.out", "w", stdout) ; 34 while (~scanf ("%d%d%d", &a, &b, &n)) 35 { 36 memset (col_hash, 0, sizeof(col_hash)) ; 37 for (i = 0 ; i < n; i++) 38 scanf ("%d%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i], &color[i]) ; 39 col_hash[1] += dfs (0,0,a,b,0) ; 40 for (i = 0 ; i < n; i++) 41 col_hash[color[i]] += dfs (x1[i], y1[i], x2[i], y2[i], i+1) ; 42 43 for (i = 1 ; i <= 2500 ; i++) if (col_hash[i] != 0) 44 printf ("%d %d ", i, col_hash[i]) ; 45 } 46 return 0 ; 47 }
Contact：暴力题。各种犯2，算错上界，看漏条件，4A。。。
1 # include <stdio.h> 2 # include <string.h> 3 # include <stdlib.h> 4 5 6 typedef struct NODE{ 7 int num, val ; 8 }NODE ; 9 10 11 char str[200010] ; 12 int dp[70000] ; 13 NODE ans[70000] ; 14 15 16 void Add(int len, int num){dp[(len<<12) + num] ++ ;} 17 int cmp(const void *a, const void *b) 18 { 19 NODE *p = (NODE*)a, *q = (NODE*)b ; 20 if (p->num != q->num) return q->num - p->num ; 21 return p->val - q->val ; 22 } 23 24 25 void prt(int val) 26 { 27 int i, len = val>>12 ; 28 val &= ((1<<12)-1) ; 29 for (i = len-1 ; i >= 0 ; i--) 30 printf ("%d", (val>>i) & 1) ; 31 } 32 33 34 void output(int n) 35 { 36 int i, j, cc ; 37 int ccc = 0 ; 38 for (i = 0, j=0 ; i < 54000 ; i++) if (dp[i] != 0) 39 { 40 ans[j].num = dp[i] ; 41 ans[j].val = i ; 42 j++ ; 43 } 44 cc = j ; 45 qsort(ans, cc, sizeof(ans[0]), cmp) ; 46 for (i = 0, j = 0 ; j < cc ; j++) 47 { 48 if (j == 0) 49 { 50 printf ("%d ", ans[j].num) ; 51 prt(ans[j].val) ; 52 ccc = 1 ; 53 } 54 else if (ans[j].num != ans[j-1].num) 55 { 56 i++ ; 57 if (i >= n) break ; 58 printf (" %d ", ans[j].num) ; 59 prt(ans[j].val) ; 60 ccc = 1; 61 } 62 else{ 63 if (ccc == 6) printf (" "), ccc = 0 ; 64 else printf (" ") ; 65 prt(ans[j].val) ; 66 ccc++ ; 67 } 68 } 69 } 70 71 72 int main () 73 { 74 int a, b, len, num, n, j, slen ; 75 freopen ("contact.in", "r", stdin) ; 76 freopen ("contact.out", "w", stdout) ; 77 scanf ("%d%d%d", &a, &b, &n) ; 78 str[0] = ' ' ; 79 while (~scanf ("%s", str+strlen(str))) ; 80 slen = strlen(str) ; 81 { 82 memset (dp, 0, sizeof(dp)) ; 83 for (len = a ; len <= b && len <= slen ; len++) 84 { 85 for (j = 0, num = 0 ; j < len ; j++) 86 num = num*2 + str[j]-'0' ; 87 Add(len, num) ; 88 for (j = len ; str[j] ; j++) 89 { 90 num = num*2 + str[j]-'0' ; 91 num &= (1<<len)-1 ; 92 Add(len, num) ; 93 } 94 } 95 output(n) ; 96 printf (" ") ; 97 } 98 return 0 ; 99 }
Stamps：简单dp。
1 # include <stdio.h> 2 3 4 int v[55] ; 5 int dp[2000010] ; 6 7 8 int max(int a, int b){return a>b?a:b;} 9 int min(int a, int b){return a<b?a:b;} 10 11 12 int main () 13 { 14 int max_v, k, n, i, j ; 15 freopen ("stamps.in", "r", stdin) ; 16 freopen ("stamps.out", "w", stdout) ; 17 while (~scanf ("%d%d", &k, &n)) 18 { 19 max_v = 0 ; 20 for (i = 0 ; i < n ; i++) 21 scanf ("%d", &v[i]), max_v = max(max_v, v[i]*k) ; 22 for (i = 1 ; i <= max_v+1 ; i++) 23 { 24 dp[i] = k+1 ; 25 for (j = 0 ; j < n ; j++) 26 { 27 if (i-v[j] >= 0) 28 dp[i] = min(dp[i], dp[i-v[j]]+1) ; 29 } 30 if (dp[i] > k) break ; 31 } 32 printf ("%d ", i-1) ; 33 } 34 return 0 ; 35 }