1 f[i][v]=( f[i-1][v] + f[i-1][v-c[i]]+w[i] ); 2 3 f[v]=max( f[v] , f[v-c[i]]+w[i] ); 4 5 f[i][v]=max( f[i-1][v], f[i-1][v-c[i]*k]+w[i]*k); 6 0<=k>=v/c[i]; 7 8 f[i][v]=max( f[i-1][v] , f[i][v-c[i]]+w[i] ); 9 10 for(int v=0;v<V;v++) 11 12 13 14 15 16 17 for(int i=0;i<N;i++) 18 for( int v=0;v<V;v++ ) 19 f[i][v]=max(f[i-1][v],f[i-1][v-c[i]]+w[i]); 20 21 22 for(int i=0;i<N;i++) 23 for( int v=V;v>=0;v--) 24 f[v]=max(f[v],f[V-c[i]]+w[i]); 25 26 for(int i=0;i<N;i++) 27 for(int v=0;v<V;v++) 28 f[i][v]=max(f[i-1][v],f[i][V-c[i]]+w[i]); 29 30 for(int i=0;i<N;i++) 31 for(int v=0;v<V;v++) 32 f[v]=max(f[v],f[V-c[i]]+w[i]); 33 34 35 36 37 38 //01背包 39 for（int i = 1;i< = n; i++） 40 for(int j = V; j >= a[i] ; j--) 41 f[j] = max ( f[j], f[ j - a[i] ] + w[i] ) 42 43 44 //完全背包 45 for( int i = 1;i <= n; i++) 46 for( int j = a[i];j <= V; j++) 47 f[j] = max( f[j] , f[ j - a[i] ] + w[i] );
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背包记录路径
https://www.patest.cn/contests/gplt/L3-001
找零钱
1 #include <iostream> 2 #include <algorithm> 3 #include <string.h> 4 using namespace std; 5 6 const int maxn = 1e4+10; 7 int n, m, a[maxn], b[maxn], cnt[maxn], pr[maxn]; 8 9 void pri(int sum){ 10 if(sum != cnt[sum]){ 11 pri(sum - cnt[sum]); 12 printf(" %d", cnt[sum]); 13 }else{ 14 printf("%d", cnt[sum]); 15 } 16 } 17 18 int main(){ 19 scanf("%d%d", &n, &m); 20 for(int i = 1; i <= n; ++i){ 21 scanf("%d", &a[i]); 22 } 23 memset(b, -0x3f3f, sizeof(b)); 24 sort(a+1, a+n+1); 25 b[0] = 0; 26 for(int i = 1; i <= n; ++i){ 27 for(int j = m; j >= a[i]; --j){ 28 if(b[j] <= b[j-a[i]]+1) b[j] = b[j-a[i]]+1, cnt[j] = a[i]; 29 } 30 } 31 if(b[m] > 0){ 32 pri(m); 33 printf(" "); 34 }else{ 35 printf("No Solution "); 36 } 37 return 0; 38 }
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多重背包记录路径
http://poj.org/problem?id=1787
1 #include <iostream> 2 #include <string.h> 3 #include <stdio.h> 4 using namespace std; 5 6 const int maxn = 10009; 7 int b[4], e[4], d[maxn], c[maxn], us[maxn]; 8 9 int main() { 10 int p, a[] = {25, 10, 5, 1}; 11 while(~scanf("%d%d%d%d%d", &p, &b[3], &b[2], &b[1], &b[0])) { 12 if(p == 0 && b[3] == 0 && b[2] == 0 && b[1] == 0 && b[0] == 0) break; 13 memset(d, -1, sizeof(d)); 14 d[0] = 0; 15 for(int i = 3; i >= 0; --i) { 16 memset(us, 0, sizeof(us)); 17 for(int j = a[i]; j <= p; ++j) { 18 if(d[j-a[i]] != -1 && d[j] <= d[j-a[i]]+1 && us[j-a[i]] < b[i]) { 19 d[j] = d[j-a[i]]+1; 20 us[j] = us[j-a[i]] + 1; 21 c[j] = i; 22 } 23 } 24 } 25 //cout<<d[p]<<endl; 26 if(d[p] > 0) { 27 e[0] = e[1] = e[2] = e[3] = 0; 28 while(d[p]) { 29 e[c[p]]++; 30 p -= a[c[p]]; 31 } 32 printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters. ", e[3], e[2], e[1], e[0]); 33 34 } else { 35 printf("Charlie cannot buy coffee. "); 36 } 37 } 38 return 0; 39 }
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01背包记录路径
http://codeforces.com/contest/864/problem/E
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 5 const int maxn = 111; 6 int a[maxn], b[maxn], c[maxn], id[maxn], d[111][2009], pre[111][2009], num = 0; 7 8 void pri(int xx, int mx){ 9 if(xx == 0){ 10 printf("%d ", num); 11 return ; 12 } 13 if(pre[xx][mx] != mx) num ++; 14 pri(xx-1, pre[xx][mx]); 15 if(pre[xx][mx] != mx){ 16 printf("%d ", id[xx]); 17 } 18 } 19 20 int main(){ 21 int n, mx = 0, ma = 0, xx; 22 scanf("%d", &n); 23 for(int i = 1; i <= n; ++i){ 24 scanf("%d%d%d", &a[i], &b[i], &c[i]); 25 id[i] = i; 26 } 27 sort(id+1, id+n+1, [](int x, int y){ 28 return b[x] < b[y]; 29 }); 30 int sum = 0; 31 for(int i = 1; i <= n; ++i){ 32 int x = id[i]; 33 for(int k = 0; k < 2009; ++k) d[i][k] = d[i-1][k], pre[i][k] = k; 34 for(int j = b[x]; j > a[x]; --j){ 35 if(d[i-1][j-a[x]]+c[x] > d[i][j]){ 36 d[i][j] = d[i-1][j-a[x]] + c[x]; 37 pre[i][j] = j-a[x]; 38 } 39 } 40 } 41 for(int i = 0; i < 2009; ++i){ 42 if(d[n][i] > ma) ma = d[n][i], mx = i; 43 } 44 printf("%d ", ma); 45 pri(n, mx); 46 printf(" "); 47 return 0; 48 }
大牛代码
1 #include <vector> 2 #include <iostream> 3 #include <algorithm> 4 using namespace std; 5 int n, t[109], d[109], p[109], id[109], dp[109][2009]; 6 int main() { 7 cin >> n; 8 for (int i = 0; i < n; i++) cin >> t[i] >> d[i] >> p[i], id[i] = i + 1; 9 for (int i = 0; i < n; i++) { 10 for (int j = i + 1; j < n; j++) { 11 if (d[i] > d[j]) { 12 swap(t[i], t[j]); 13 swap(d[i], d[j]); 14 swap(p[i], p[j]); 15 swap(id[i], id[j]); 16 } 17 } 18 } 19 for (int i = 0; i < n; i++) { 20 for (int j = 0; j < d[i]; j++) { 21 dp[i + 1][j] = dp[i][j]; 22 if (j >= t[i]) dp[i + 1][j] = max(dp[i + 1][j], dp[i][j - t[i]] + p[i]); 23 } 24 } 25 int pos = max_element(dp[n], dp[n] + d[n - 1]) - dp[n]; 26 cout << dp[n][pos] << endl; 27 vector<int> v; 28 for (int i = n - 1; i >= 0; i--) { 29 if (dp[i][pos] != dp[i + 1][pos]) v.push_back(id[i]), pos -= t[i]; 30 } 31 reverse(v.begin(), v.end()); 32 cout << v.size() << endl; 33 for (int i = 0; i < v.size(); i++) cout << v[i] << ' '; 34 return 0; 35 }
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