HDU1789 Doing Homework again 【贪心】

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6878    Accepted Submission(s): 4096

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

 

Input

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

 

Output

For each test case, you should output the smallest total reduced score, one line per test case.

 

Sample Input

3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

 

Sample Output

0
3
5

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

题意：有n个任务，每一个任务完毕须要一整天。给出每一个任务完毕的期限和完不成的惩处值。求最小惩处值。

题解：看完这题的第一反应是二分图最优匹配。然后就直接上模板了。结果是超时，后来发现这题能够用贪心，每次选择惩处值最大的任务安排，安排的规则为从该任务的期限后面往前选择一个没有安排任务的天数，将该任务安插在这一天。

贪心：

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 1010

struct Node {
	int days, penalty;
} arr[maxn];
bool vis[maxn];

bool cmp(Node a, Node b) {
	return a.penalty > b.penalty;
}

int main() {
	// freopen("stdin.txt", "r", stdin);
	int t, n, i, j, sum, sum2;
	scanf("%d", &t);
	while(t--) {
		scanf("%d", &n);
		sum = sum2 = 0;
		for(i = 0; i < n; ++i) 
			scanf("%d", &arr[i].days);
		for(i = 0; i < n; ++i) {
			scanf("%d", &arr[i].penalty);
			sum += arr[i].penalty;
		}
		std::sort(arr, arr + n, cmp);
		memset(vis, 0, sizeof(vis));
		for(j = 0; j < n; ++j)
			for(i = arr[j].days; i; --i)
				if(!vis[i]) {
					vis[i] = true;
					sum2 += arr[j].penalty;
					break;
				}
		printf("%d
", sum - sum2);
	}
	return 0;
}

TLE的KM：

#include <stdio.h>
#include <string.h>

#define maxn 1010
#define inf 0x3f3f3f3f

int G[maxn][maxn];
int Lx[maxn], Ly[maxn];
int match[maxn];
bool visx[maxn], visy[maxn];
int slack[maxn], n, m, sum;

bool DFS(int cur) {
	int t, y;
	visx[cur] = true;
	for(y = 1; y <= n; ++y) {
		if(visy[y]) continue;
		t = Lx[cur] + Ly[y] - G[cur][y];
		if(0 == t) {
			visy[y] = true;
			if(-1 == match[y] || DFS(match[y])) {
				match[y] = cur; return true;
			}
		} else if(slack[y] > t) slack[y] = t;
	}
	return false;
}

int KM() {
	int i, j, x, d, ret = 0;
	memset(match, -1, sizeof(int) * (n + 1));
	memset(Ly, 0, sizeof(int) * (n + 1));
	for(i = 1; i <= n; ++i) {
		Lx[i] = -inf;
		for(j = 1; j <= n; ++j)
			if(G[i][j] > Lx[i]) Lx[i] = G[i][j];
	}
	for(x = 1; x <= n; ++x) {
		memset(slack, 0x3f, sizeof(int) * (n + 1));
		while(true) {
			memset(visx, 0, sizeof(bool) * (n + 1));
			memset(visy, 0, sizeof(bool) * (n + 1));
			if(DFS(x)) break;
			d = inf;
			for(i = 1; i <= n; ++i)
				if(!visy[i] && d > slack[i])
					d = slack[i];
			for(i = 1; i <= n; ++i)
				if(visx[i]) Lx[i] -= d;
			for(i = 1; i <= n; ++i)
				if(visy[i]) Ly[i] += d;
				else slack[i] -= d;
		}
	}
	for(i = 1; i <= n; ++i)
		if(match[i] > -1) 
			ret += G[match[i]][i];
	return ret;
}

void getMap() {
	int i, w, j;
	sum = 0;
	scanf("%d", &n);
	memset(G, 0, sizeof(G));
	for(i = 1; i <= n; ++i)
		scanf("%d", &match[i]); // 暂时表示第i个任务能够匹配的天数
	for(i = 1; i <= n; ++i) {
		scanf("%d", &w);
		sum += w;
		for(j = 1; j <= match[i]; ++j)
			G[i][j] = w;
	}
}

void solve() {
	printf("%d
", sum - KM());
}

int main()
{
	// freopen("stdin.txt", "r", stdin);
	int t;
	scanf("%d", &t);
	while(t--) {
		getMap();
		solve();
	}
}