POJ 1949 Chores(DP)
2015-01-14 15:33
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题意:n个任务,每个任务有一个完成时间,每个任务有一些前趋任务,问最短需要多少时间可以完成全部任务,(任务可以同时执行,但是一个任务必须在他前趋任务都完成了之后才可以去完成)
思路:DAG上DP,每个任务的完成时间就是dp[u] = min(dp[v] + time),v是u的子结点
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 10005;
int n, ti
, dp
;
vector<int> g
;
int dfs(int u) {
if (dp[u] != -1) return dp[u];
dp[u] = ti[u];
int Max = 0;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
Max = max(Max, dfs(v));
}
dp[u] += Max;
return dp[u];
}
int main() {
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; i++) {
g[i].clear();
scanf("%d", &ti[i]);
int tot;
scanf("%d", &tot);
while (tot--) {
int v;
scanf("%d", &v);
g[i].push_back(v);
}
}
int ans = 0;
memset(dp, -1, sizeof(dp));
for (int i = 1; i <= n; i++)
ans = max(ans, dfs(i));
printf("%d\n", ans);
}
return 0;
}
思路:DAG上DP,每个任务的完成时间就是dp[u] = min(dp[v] + time),v是u的子结点
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 10005;
int n, ti
, dp
;
vector<int> g
;
int dfs(int u) {
if (dp[u] != -1) return dp[u];
dp[u] = ti[u];
int Max = 0;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
Max = max(Max, dfs(v));
}
dp[u] += Max;
return dp[u];
}
int main() {
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; i++) {
g[i].clear();
scanf("%d", &ti[i]);
int tot;
scanf("%d", &tot);
while (tot--) {
int v;
scanf("%d", &v);
g[i].push_back(v);
}
}
int ans = 0;
memset(dp, -1, sizeof(dp));
for (int i = 1; i <= n; i++)
ans = max(ans, dfs(i));
printf("%d\n", ans);
}
return 0;
}
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