編程演算法 - 背包問題(三種動態規劃) 代碼(C)

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背包問題(三種動態規劃) 代碼(C)

本文地址: http://blog.csdn.net/caroline_wendy

題目參考: http://blog.csdn.net/caroline_wendy/article/details/37912949

可以用動態規劃(Dynamic Programming, DP)求解, 可以通過記憶化搜尋推匯出遞推式, 可以使用三種不同的方向進行求解.

動態規劃主要是狀態轉移, 需要理解清晰.

代碼:

/* * main.cpp * *  Created on: 2014.7.17 *      Author: spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <memory.h>#include <limits.h>#include <utility>#include <queue>#include <algorithm>using namespace std;class Program {static const int MAX_N = 100;int n=4, W=5;int w[MAX_N] = {2,1,3,2}, v[MAX_N]={3,2,4,2};int dp[MAX_N+1][MAX_N+1]; //預設初始化為0public:void solve() {for (int i=n-1; i>=0; i--) {for (int j=0; j<=W; ++j) {if (j<w[i]) {dp[i][j] = dp[i+1][j];} else {dp[i][j] = max(dp[i+1][j], dp[i+1][j-w[i]] + v[i]);}}}printf("result = %d\n", dp[0][W]);}void solve1() {for (int i=0; i<n; ++i) {for (int j=0; j<=W; ++j) {if (j<w[i]) {dp[i+1][j] = dp[i][j];} else {dp[i+1][j] = max(dp[i][j], dp[i][j-w[i]]+v[i]);}}}printf("result = %d\n", dp[n][W]);}void solve2() {for (int i=0; i<n; i++) {for (int j=0; j<=W; ++j) {dp[i+1][j] = max(dp[i+1][j], dp[i][j]);if (j+w[i]<=W) {dp[i+1][j+w[i]] = max(dp[i+1][j+w[i]], dp[i][j]+v[i]);}}}printf("result = %d\n", dp[n][W]);}};int main(void){Program P;P.solve2();    return 0;}

輸出:

result = 7

節省空間的, 可以使用1維數組的動態規劃.

代碼:

/* * main.cpp * *  Created on: 2014.7.17 *      Author: spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <memory.h>#include <limits.h>#include <utility>#include <queue>#include <algorithm>using namespace std;class Program {static const int MAX_N = 100;int n=4, W=5;int w[MAX_N] = {2,1,3,2}, v[MAX_N]={3,2,4,2};int dp[MAX_N+1];public:void solve() {memset(dp, 0, sizeof(dp));for (int i=0; i<n; ++i) {for (int j=W; j>=w[i]; --j) {dp[j] = max(dp[j], dp[j-w[i]]+v[i]);}}printf("result = %d\n", dp[W]);}};int main(void){Program P;P.solve();    return 0;}

輸出:

result = 7