动态规划：背包问题

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

/*
4 5
2 3 1 2 3 4 2 2
*/
const int maxn = 100 + 20;
const int maxv = 100 + 20;
const int INF = 100000000;
int w[maxn], v[maxn];
int N, W;
int dp[maxn + 1][maxv*maxn + 100];   //

int dfs(int i, int j)
{
    if (dp[i][j] >= 0) {
        return dp[i][j];
    }    
    int res;
    if (i == N) {
        res = 0;
    } else if (j < w[i]) {
        res = dfs(i + 1, j);
    } else {
        res = max(dfs(i + 1, j), dfs(i + 1, j - w[i]) + v[i]);
    }
    return dp[i][j] = res;
}

void solve()
{
    cin >> N >> W;
    for (int i = 0; i < N; i++) {
        cin >> w[i] >> v[i];
    }    
    memset(dp, -1, sizeof(dp));
    int res = dfs(0, W);
    cout << res << endl;
}

void solve1()
{
    cin >> N >> W;
    for (int i = 0; i < N; i++) {
        cin >> w[i] >> v[i];
    }
    memset(dp, 0, sizeof(dp));
    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("%d
", dp[0][W]);
}

void solve2()
{
    cin >> N >> W;
    for (int i = 0; i < N; i++) {
        cin >> w[i] >> v[i];
    }
    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= W; j++) {
            if (j < w[j]) {
                dp[i + 1][j] = dp[i][j];
            } else {
                dp[i + 1][j] = max(dp[i][j], dp[i][j - w[i]] + v[i]);
            }
        }
    }
    printf("%d
", dp[N][W] + 1);
}

void solve3()
{
    cin >> N >> W;
    for (int i = 0; i < N; i++) {
        cin >> w[i] >> v[i];
    }
    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("%d
", dp[N][W]);
}

//大数, wi <= 10^7 
void solve4()
{
    //前i-1个物品中挑选出价值总和为j 的部分
    cin >> N >> W;
    for (int i = 0; i < N; i++) {
        cin >> w[i] >> v[i];
    }
    
    fill(dp[0], dp[0] + maxn*maxv + 1, INF);
    dp[0][0] = 0;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= maxn * maxv + 1; j++) {
            if (j < v[i]) {         //挑选V价值的最小重量
                dp[i + 1][j] = dp[i][j];
            } else {
                dp[i + 1][j] = min(dp[i][j], dp[i][j - v[i]] + w[i]);
            }
        }
    }
    
    int res = 0;
    for (int i = 0; i <= maxn * maxv; i++) {
        if (dp[N][i] <= W) res = i;        //可以得到最大value的那个下标
    }
     printf("%d
", res);
}

int main()
{
//    solve();
//     
//    solve1();
    
//    solve2();

//    solve3();

    solve4();
    
    return 0;
}