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

 

#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\n", sum - sum2);

	}

	return 0;

}

#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\n", sum - KM());

}

int main()

{

	// freopen("stdin.txt", "r", stdin);

	int t;

	scanf("%d", &t);

	while(t--) {

		getMap();

		solve();

	}

}