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
