POJ1015 Jury Compromise

柔光的暖阳◎ 2021-11-27 05:14 166阅读 0赞

## 题意：《算法竞赛进阶指南》P278。 ##

## 分析：《算法竞赛进阶指南》P278-279。 ##

## 代码： ##

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    using namespace std;
    int f[25][805]; // 前i个人，选了j个，辩方总分与控方总分的差为k=-400~400（平移到0~800）
    int d[205][25][805]; // 记路径还是要三维，不然后面的更新可能会覆盖掉之前的记录
    int n, m, a[205], b[205], suma, sumb, T;
    vector<int> c;
    
    void get_path(int i, int j, int k) {
    	if (j == 0) return;
    	int last = d[i][j][k];
    	get_path(last - 1, j - 1, k - (a[last] - b[last]));
    	c.push_back(last);
    	suma += a[last], sumb += b[last];
    }
    
    int main() {
    	while (cin >> n >> m && n) {
    		for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
    		memset(f, 0xcf, sizeof(f)); // -INF
    		f[0][400] = 0; // f[0][0]第三维平移400
    		for (int i = 1; i <= n; i++) {
    			for (int j = 0; j <= m; j++) // 不选i
    				for (int k = 0; k <= 800; k++) d[i][j][k] = d[i - 1][j][k];
    			for (int j = m; j; j--) // 选i
    				for (int k = 0; k <= 800; k++) {
    					if (k - (a[i] - b[i]) < 0 || k - (a[i] - b[i]) > 800) continue; // 超出范围
    					if (f[j][k] < f[j - 1][k - (a[i] - b[i])] + a[i] + b[i]) {
    						f[j][k] = f[j - 1][k - (a[i] - b[i])] + a[i] + b[i];
    						d[i][j][k] = i;
    					}
    				}
    		}
    		int ans = 0;
    		for (int k = 0; k <= 400; k++) {
    			if (f[m][400 + k] >= 0 && f[m][400 + k] >= f[m][400 - k]) {
    				ans = k + 400;
    				break;
    			}
    			if (f[m][400 - k] >= 0) {
    				ans = 400 - k;
    				break;
    			}
    		}
    		c.clear();
    		suma = sumb = 0;
    		get_path(n, m, ans);
    		printf("Jury #%d\n", ++T);
    		printf("Best jury has value %d for prosecution and value %d for defence:\n", suma, sumb);
    		for (int i = 0; i < c.size(); i++) printf(" %d", c[i]);
    		printf("\n\n");
    	}
    }