POJ 1015 Jury Compromise——01背包变形

这是uva323的题解，把这个题的数据加强了

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>using namespace std;const int maxn = 210;int kase, n, m, d, p, add[maxn], sub[maxn], dp[30][maxn<<2];vector<int> path[30][maxn<<2];int main() {    while (~scanf("%d %d", &n, &m) && n && m) {        for (int i = 1; i <= n; i++) {            scanf("%d %d", &d, &p);            add[i] = d + p, sub[i] = d - p;        }        int fix = 20 * m;        memset(dp, -1, sizeof(dp)); dp[0][fix] = 0;        for (int i = 1; i <= n; i++) {            for (int j = m; j >= 1; j--) {                for (int k = 0; k < 2 * fix; k++) {                    if (0 > k + sub[i] || k + sub[i] > 2 * fix) continue;                    if (dp[j - 1][k] == -1) continue;                    if (dp[j][k+sub[i]] < dp[j - 1][k] + add[i]) {                        dp[j][k+sub[i]] = dp[j - 1][k] + add[i];                        path[j][k+sub[i]] = path[j - 1][k];                        path[j][k+sub[i]].push_back(i);                    }                }            }        }        int ans = 0;        while (dp[m][fix - ans] == -1 && dp[m][fix + ans] == -1) ans++;        int temp = (dp[m][fix + ans] > dp[m][fix - ans]) ? fix + ans : fix - ans;        printf("Jury #%d\nBest jury has value %d for prosecution and value %d for defence:\n",               ++kase, (dp[m][temp] + temp - fix) / 2, (dp[m][temp] - temp + fix) / 2);        for (int i = 0; i < m; i++) {            printf(" %d", path[m][temp][i]);        }        printf("\n\n");    }    return 0;}