POJ 1015 Jury Compromise（DP）

来源：互联网发布：ember.js 开发工具编辑：程序博客网时间：2024/06/03 11:17

题意：每个人有a， b两个值，现在从n个人选出m个人，使得a和b的总和之差绝对值最小，如果有多解，输出a和b总和的和最大的解

思路：dp[i][j][k]表明选到第i个人，选了j个人，差为k的解的总和最大值，然后转移最后找一下解即可

代码：

#include <cstdio>#include <cstring>#include <cstdlib>const int N = 205;const int M = 25;const int INF = 0x3f3f3f3f;int n, m, d[N], p[N], dp[N][M][805], path[N][M][805], out[M], on;int main() {    int cas = 0;    while (~scanf("%d%d", &n, &m) && n || m) {for (int i = 0; i <= n; i++)    for (int j = 0; j <= m; j++)for (int k = 0; k <= 800; k++)    dp[i][j][k] = -INF;dp[0][0][400] = 0;for (int i = 1; i <= n; i++) {    scanf("%d%d", &d[i], &p[i]);    for (int j = 0; j <=m; j++) {for (int k = 0; k <= 800; k++) {    if (dp[i][j][k] < dp[i - 1][j][k]) {dp[i][j][k] = dp[i - 1][j][k];path[i][j][k] = 0;    }    if (j == 0) continue;    int pre = k - d[i] + p[i];    if (pre < 0 || pre > 800 || dp[i - 1][j - 1][pre] == -INF) continue;    if (dp[i][j][k] < dp[i - 1][j - 1][pre] + d[i] + p[i]) {dp[i][j][k] = dp[i - 1][j - 1][pre] + d[i] + p[i];path[i][j][k] = i;    }}    }}int ans = INF, ansv;for (int i = 0; i <= 800; i++) {    if (dp[n][m][i] != -INF && ans > abs(i - 400)) {ans = abs(i - 400);ansv = i;    } else if (ans == abs(i - 400)) {if (dp[n][m][i] > dp[n][m][ansv])    ansv = i;    }}int ans1 = 0, ans2 = 0;on = 0;while (n) {    int tmp = path[n][m][ansv];    if (tmp) {out[on++] = tmp;n--;m--;ansv = ansv - d[tmp] + p[tmp];ans1 += d[tmp];ans2 += p[tmp];    } elsen--;}printf("Jury #%d\n", ++cas);printf("Best jury has value %d for prosecution and value %d for defence:\n", ans1, ans2);for (int i = on - 1; i >= 0; i--)    printf(" %d", out[i]);printf("\n\n");    }    return 0;}

0 0