HDU1789 Doing Homework again 【贪心】
Doing Homework again
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.
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.
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
0
3
5
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\n", 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\n", sum - KM());
} int main()
{
// freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--) {
getMap();
solve();
}
}
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