五项一起对得上吗

运行下面这段程序：

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
using Item = pair<int, int>;   // (重量, 价值)

const vector<Item> ITEMS = {{3, 8}, {4, 9}, {5, 11}};

vector<long long> knap1d(const vector<Item>& items, int cap) {
    vector<long long> dp(cap + 1, 0);
    for (auto [w, v] : items) {
        for (int c = cap; c >= w; c--) {
            if (dp[c - w] + v > dp[c]) dp[c] = dp[c - w] + v;
        }
    }
    return dp;
}

vector<long long> knap_full(const vector<Item>& items, int cap) {
    vector<long long> dp(cap + 1, 0);
    for (auto [w, v] : items) {
        for (int c = w; c <= cap; c++) {
            if (dp[c - w] + v > dp[c]) dp[c] = dp[c - w] + v;
        }
    }
    return dp;
}

long long merge_cost(const vector<int>& a) {
    int n = a.size();
    const long long INF = 1000000000000000000LL;
    vector<long long> pre(n + 1, 0);
    for (int i = 0; i < n; i++) pre[i + 1] = pre[i] + a[i];
    vector<vector<long long>> dp(n, vector<long long>(n, 0));
    for (int L = 2; L <= n; L++) {
        for (int i = 0; i + L - 1 < n; i++) {
            int j = i + L - 1;
            dp[i][j] = INF;
            for (int k = i; k < j; k++) {
                long long t = dp[i][k] + dp[k + 1][j] + pre[j + 1] - pre[i];
                if (t < dp[i][j]) dp[i][j] = t;
            }
        }
    }
    return dp[0][n - 1];
}

long long paths(int m, int n) {
    vector<vector<long long>> dp(m, vector<long long>(n, 1));
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
    }
    return dp[m - 1][n - 1];
}

int edit(const string& a, const string& b) {
    int m = a.size(), n = b.size();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    for (int i = 0; i <= m; i++) dp[i][0] = i;
    for (int j = 0; j <= n; j++) dp[0][j] = j;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1];
            else dp[i][j] = 1 + min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]});
        }
    }
    return dp[m][n];
}

int main() {
    bool ok = knap1d(ITEMS, 10)[10] == 20 && knap_full(ITEMS, 10)[10] == 25
        && merge_cost({4, 1, 2, 3}) == 19 && paths(3, 4) == 10 && edit("horse", "ros") == 3;
    cout << boolalpha << ok << endl;
}

（本题用 g++ -std=c++17 -O0 编译。）