题目大意：john有H个小时，他要去钓鱼。已知有N个池塘，排在一起，池塘间有间隔，从i个池塘走到i+1需要时间ti。john必须从第一个池塘开始，往2-3-4...池塘的顺序钓鱼。每个池塘刚开始每分钟调fi与，往后递减。。。求最多调多少鱼。。
我的思路：dp
f[i][j]表示前i个池塘用j分钟所能钓到最多的鱼。。
f[i][j] = max(f[i - 1][k] + a[i][j - k - ti]);
a[i][j-k-ti] 表示第i个池塘给定j-k-ti分钟所能钓到的鱼、、、
。。输出比较蛋疼。。
还有，可知5分钟为一个单位、、先除5再说
1 /* 2 State:Accepted 3 Time:2013.2.28 4 */ 5 #include <iostream> 6 #include <cstdlib> 7 #include <cstring> 8 #include <string> 9 #include <cstdio> 10 #include <cmath> 11 #include <algorithm> 12 #include <vector> 13 #define minn -1000000; 14 using namespace std; 15 int h , n , first[100], decl[100] ,dt[100] , f[30][200] , a[30][200] , value[30][200]; 16 int map[30][200] , Time[30][200], ans , ansi, ansj , result[30]; 17 18 void init(){ 19 scanf("%d",&h); 20 h *= 12; 21 int temp ,tm; 22 memset(a , 0 ,sizeof(a)); 23 memset(value, 0 , sizeof(value)); 24 memset(result, 0 ,sizeof(result)); 25 memset(map , 0 , sizeof(map)); 26 for (int i = 1; i <= n; ++i) 27 scanf("%d",&first[i]); 28 for (int i = 1; i <= n; ++i) 29 scanf("%d",&decl[i]); 30 for (int i = 1; i <= n - 1; ++i) 31 scanf("%d",&dt[i]); 32 for (int i = 1; i <= n; ++i){ 33 temp = tm = first[i]; 34 a[i][0] = dt[i - 1];; 35 for (int j = 1; j <= h; ++j){ 36 a[i][j] = j + dt[i - 1]; 37 value[i][j] = temp; 38 tm -= decl[i]; 39 if (tm > 0) temp += tm; 40 } 41 } 42 } 43 44 bool footprint(int i , int opt1 , int opt2){ 45 if (i == 1 || i == 0){ 46 if ( Time[i][opt1] > Time[i][opt2] ) return true; 47 return false; 48 } 49 bool bo = footprint(i - 1, map[i][opt1], map[i][opt2]); 50 if ( bo || Time[i][opt1] > Time[i][opt2] ) return true; 51 return false; 52 } 53 54 int findjp(int lasti, int i , int opt){ 55 if (i == lasti) return opt; 56 return findjp(lasti , i - 1, map[i][opt]); 57 } 58 59 void dp(){ 60 int tp ,tv , nowj; 61 ans = 0; 62 bool bo; 63 for (int i = 0; i <= n; ++i) 64 for (int j = 0; j <= h; ++j) 65 f[i][j]= minn; 66 ans = f[0][0] = 0; 67 68 for (int i = 1; i <= n; ++i) 69 for (int j = 0; j <= h; ++j) 70 for (int k = 0 ; k <= h; ++k){ 71 if (j - a[i][k] < 0) break; 72 tp = j - a[i][k]; 73 tv = value[i][k]; 74 if (f[i - 1][tp] + tv > f[i][j] 75 || f[i - 1][tp] +tv == f[i][j] && 76 footprint(i - 1 , tp , map[i][j]) ){ 77 f[i][j] = f[i-1][tp] + tv; 78 map[i][j] = tp; 79 Time[i][j] = k; 80 } 81 } 82 for (int i = 1; i <= n; ++i) 83 for (int j = 1; j <= h; ++j){ 84 bo = false; 85 if (f[i][j] > ans) bo =true; 86 if (f[i][j] == ans){ 87 if (ansi >= i) nowj = findjp(i , ansi ,ansj); 88 if (ansi < i) nowj = findjp(ansi , i , j); 89 if (ansi >= i && footprint(i , j , nowj)) bo =true; 90 if (ansi < i && footprint(ansi , nowj ,ansj)) bo = true; 91 } 92 if (bo){ 93 ans = f[i][j]; 94 ansi = i; 95 ansj = j; 96 } 97 } 98 99 } 100 101 void find(int i , int j){ 102 if (i == 0) return; 103 result[i] = Time[i][j] * 5; 104 find(i - 1, map[i][j]); 105 } 106 107 void print(){ 108 find(ansi ,ansj); 109 if (ans == 0) result[1] = h * 5; 110 for (int i = 1; i <= n - 1; ++i) 111 printf("%d, " , result[i]); 112 printf("%d\n",result[n]); 113 printf("Number of fish expected: %d\n\n", ans); 114 115 } 116 117 int main(){ 118 freopen("poj1042.in","r",stdin); 119 freopen("poj1042.out","w",stdout); 120 scanf("%d",&n); 121 while (n){ 122 init(); 123 dp(); 124 print(); 125 scanf("%d",&n); 126 } 127 fclose(stdin); fclose(stdout); 128 }