P1064-金明的预算方案

 1 #include <bits/stdc++.h>
 2 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 3 typedef long long ll;
 4 using namespace std;
 5 inline ll read()
 6 {
 7     ll ans = 0;
 8     char ch = getchar(), last = ' ';
 9     while(!isdigit(ch)) last = ch, ch = getchar();
10     while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
11     if(last == '-') ans = -ans;
12     return ans;
13 }
14 inline void write(ll x)
15 {
16     if(x < 0) x = -x, putchar('-');
17     if(x >= 10) write(x / 10);
18     putchar(x % 10 + '0');
19 }
20 int N,m;
21 int rs[3][60];
22 struct g
23 {
24     int n;
25     int pre;
26     int c[5];
27     int v[5];
28 };
29 g rl[65];
30 int search(int t)
31 {
32     _for(i,0,65)
33         if(rl[i].pre==t)
34             return i;
35     return -1;
36 }
37 
38 int rlend = 0;
39 int rnt = 0;
40 int dp[100003];
41 int main()
42 {
43     memset(dp,0,sizeof(dp));
44     N = read(),m = read();
45     _for(i,0,m)
46         _for(j,0,3)
47             rs[j][i] = read();
48     
49     _for(i,0,m)
50         if(rs[2][i]==0)
51         {
52             g tmp;
53             tmp.pre = i+1;
54             tmp.n = 1;
55             tmp.c[0] = rs[0][i];
56             tmp.v[0] = rs[1][i]*rs[0][i];
57             rl[rlend++] = tmp;
58         }
59         else
60         {
61             int binx = search(rs[2][i]);
62             if(rl[binx].n==1)
63             {
64                 rl[binx].n = 2;
65                 rl[binx].c[1] = rs[0][i]+rl[binx].c[0];
66                 rl[binx].v[1] = rs[1][i]*rs[0][i]+rl[binx].v[0];
67             }
68             else if(rl[binx].n==2)
69             {
70                 rl[binx].n = 4;
71                 rl[binx].c[2] = rs[0][i]+rl[binx].c[0];
72                 rl[binx].v[2] = rs[1][i]*rs[0][i]+rl[binx].v[0];
73                 rl[binx].c[3] = rs[0][i]+rl[binx].c[1];
74                 rl[binx].v[3] = rs[1][i]*rs[0][i]+rl[binx].v[1];
75             }
76         }
77         
78 
79     _for(i,0,rlend)
80         for(int j = N+1;j >= 0;j --)
81         {
82             _for(k,0,rl[i].n)
83                 if(j>=rl[i].c[k])
84                     dp[j] = max(dp[j],dp[j-rl[i].c[k]]+rl[i].v[k]);
85         }
86     
87     _for(i,0,100003)
88         rnt = max(rnt,dp[i]);
89     write(rnt);
90     return 0;
91 }

数据量不止一万

标签：tmp,rs,int,binx,rl,预算,P1064,dp,金明
来源： https://www.cnblogs.com/Asurudo/p/11309398.html