有M(M<=100)个气球,有N(N<=10)个人可以吹气球,但是只有一个工具,对于每个人i,有两个参数:
Ai(Ai<=10)⇒一分钟可以吹Ai个气球,Bi(Bi<=4)⇒吹完一分钟后需要休息Bi分钟才能继续吹。
问吹完M个气球最少要多少分钟。
解题思路:
观察发现Bi<=4,那么显然我们可以开一个F[sum][a1][a2][a3][a4]表示还有sum个气球,前一分钟是a1在吹,前两分钟是a2在吹,以此类推。
然后就很好做了。直接记忆化搜索就可以AC了。
AC代码:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
int F[102][11][11][11][11]={{{{{0}}}}};
int N,M;
int A[11]={0};
int B[11]={0};
int done(int sum,int a1,int a2,int a3,int a4)
{
if(sum<=0) sum=0;
if(F[sum][a1][a2][a3][a4]!=0x3f3f3f3f) return F[sum][a1][a2][a3][a4];
if(sum==0)
F[sum][a1][a2][a3][a4]=0;
else
{
int flag=0;
for(int i=1;i<=N;i++)
{
if(a1==i && B[i]>0) continue;
if(a2==i && B[i]>1) continue;
if(a3==i && B[i]>2) continue;
if(a4==i && B[i]>3) continue;
flag=1;
int gg=done(sum-A[i],i,a1,a2,a3);
F[sum][a1][a2][a3][a4]=min(F[sum][a1][a2][a3][a4],gg+1);
}
if(flag==0)
{
int gg=done(sum,0,a1,a2,a3);
F[sum][a1][a2][a3][a4]=min(F[sum][a1][a2][a3][a4],gg+1);
}
}
return F[sum][a1][a2][a3][a4];
}
int main()
{
scanf("%d%d",&M,&N);
memset(F,0x3f3f3f3f,sizeof(F));
for(int i=1;i<=N;i++)
scanf("%d%d",&A[i],&B[i]);
printf("%d",done(M,0,0,0,0));
return 0;
}
#include <iostream> #include <queue> #include <algorithm> #include <cstring> using namespace std; struct NOde { int x,y,z; } people[10]; int m,n1; queue<int>Q; struct number { int x,y,z; } num[10]; int cmp(number aa,number nn) { return aa.y > nn.y; } int cmp1(number aa,number nn) { return aa.x > nn.x; } int bfs() { for(int i = 0; i < n1; i++) { Q.push(people[i].x); Q.push(people[i].y); } int sum; int sumk = 0; while(sum < m) { int cnt = 0; memset(num,0,sizeof(num)); for(int i = 0; i < n1; i++) { int ai = Q.front(); Q.pop(); int bi = Q.front(); Q.pop(); if(people[i].z == 0) { num[i].x = ai; num[i].y = bi; num[i].z = i; cnt++; } else people[i].z--; Q.push(ai); Q.push(bi); } sort(num,num+n1,cmp); if(num[cnt-1].x!=0) { sum += num[cnt-1].x; int rr = num[cnt-1].z; people[rr].z = people[rr].y; sumk++; } else sumk++; } return sumk; } int bfs1() { while(!Q.empty()) Q.pop(); for(int i = 0; i < n1; i++) { Q.push(people[i].x); Q.push(people[i].y); } int sum=0; int sumk = 0; while(sum < m) { int cnt = 0; memset(num,0,sizeof(num)); for(int i = 0; i < n1; i++) { int ai = Q.front(); Q.pop(); int bi = Q.front(); Q.pop(); if(people[i].z == 0) { num[i].x = ai; num[i].y = bi; num[i].z = i; cnt++; } else people[i].z--; Q.push(ai); Q.push(bi); } sort(num,num+n1,cmp1); if(num[0].x!=0) { sum += num[0].x; int rr = num[0].z; people[rr].z = people[rr].y; sumk++; } else sumk++; } return sumk; } int main() { int n; while(cin >> m >> n) { n1 = 0; for(int i = 0; i<n; i++) { int rr,ee; cin>>rr>>ee; if(rr!=0) { n1++; people[i].x = rr; people[i].y = ee; people[i].z = 0; } } int sss = bfs(); for(int i = 0; i<n; i++) people[i].z = 0; int ssss = bfs1(); int min = sss; if(ssss < sss) min = ssss; cout << min << endl; } return 0; }