标签：poj3249
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 9201 | Accepted: 2080 |
Description
Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It‘s hard to have a job, since there are swelling numbers of the unemployed. So some companies often use hard tests for their recruitment.
The test is like this: starting from a source-city, you may pass through some directed roads to reach another city. Each time you reach a city, you can earn some profit or pay some fee, Let this process continue until you reach a target-city. The boss will compute the expense you spent for your trip and the profit you have just obtained. Finally, he will decide whether you can be hired.
In order to get the job, Mr.Dog managed to obtain the knowledge of the net profit Vi of all cities he may reach (a negative Vi indicates that money is spent rather than gained) and the connection between cities. A city with no roads leading to it is a source-city and a city with no roads leading to other cities is a target-city. The mission of Mr.Dog is to start from a source-city and choose a route leading to a target-city through which he can get the maximum profit.
Input
Output
Sample Input
6 5 1 2 2 3 3 4 1 2 1 3 2 4 3 4 5 6
Sample Output
7
Hint
Source
DAG：
#include <stdio.h>
#include <string.h>
#define inf 0x7fffffff
#define maxn 100002
#define maxm 1000002
int id, cost[maxn], sta[maxn];
struct Node2{
int v, first;
} head[maxn];
struct Node{
int to, next;
} E[maxm];
bool vis[maxn], in[maxn], out[maxn];
void addEdge(int u, int v)
{
E[id].to = v; E[id].next = head[u].first;
head[u].first = id++;
}
void getMap(int n, int m)
{
int i, u, v;
for(i = 1; i <= n; ++i){
scanf("%d", &head[i].v);
head[i].v = -head[i].v; in[i] = 0;
head[i].first = -1; out[i] = 0;
vis[i] = 0; cost[i] = inf;
}
for(i = id = 0; i < m; ++i){
scanf("%d%d", &u, &v); out[v] = 1;
addEdge(v, u); in[u] = 1;
}
}
void DAG(int n)
{
int i, u, v, id2 = 0, ans = inf, tmp;
for(i = 1; i <= n; ++i)
if(!in[i]){
sta[id2++] = i; vis[i] = 1;
cost[i] = head[i].v;
if(!out[i] && cost[i] < ans) ans = cost[i];
}
while(id2){
u = sta[--id2]; vis[u] = 0;
for(i = head[u].first; i != -1; i = E[i].next){
v = E[i].to; tmp = cost[u] + head[v].v;
if(tmp < cost[v]){
cost[v] = tmp;
if(!vis[v]){
vis[v] = 1; sta[id2++] = v;
}
if(!out[v] && tmp < ans) ans = tmp;
}
}
}
//if(ans < 0) ans = - ans;
printf("%d\n", -ans);
}
int main()
{
int n, m, i, u, v;
while(scanf("%d%d", &n, &m) == 2){
getMap(n, m);
DAG(n);
}
return 0;
}再放一个RE的记忆化搜索。。。：
#include <stdio.h>
#include <string.h>
#define inf 0x7fffffff
#define maxn 100002
#define maxm 1000002
int dp[maxn], id;
struct Node2{
int first, v;
} head[maxn];
struct Node{
int to, next;
} E[maxm];
bool in[maxn], out[maxn];
void addEdge(int u, int v)
{
E[id].to = v;
E[id].next = head[u].first;
head[u].first = id++;
}
void getMap(int n, int m)
{
int i, u, v;
for(i = 1; i <= n; ++i){
scanf("%d", &head[i].v);
head[i].v = -head[i].v;
in[i] = out[i] = 0; dp[i] = inf;
head[i].first = -1;
}
for(i = 0; i < m; ++i){
scanf("%d%d", &u, &v);
addEdge(u, v); in[v] = 1; out[u] = 1;
}
}
int DFS(int k)
{
if(dp[k] != inf) return dp[k];
int i, u, v, ans = inf, tmp;
for(i = head[k].first; i != -1; i = E[i].next){
tmp = head[k].v + DFS(E[i].to);
if(tmp < ans) ans = tmp;
}
return dp[k] = ans;
}
void solve(int n)
{
int i, u, v, ans = inf, tmp;
for(i = 1; i <= n; ++i)
if(!out[i]) dp[i] = head[i].v;
for(i = 1; i <= n; ++i){
if(!in[i]){
tmp = DFS(i);
if(tmp < ans) ans = tmp;
}
}
printf("%d\n", -ans);
}
int main()
{
int n, m;
while(scanf("%d%d", &n, &m) == 2){
getMap(n, m);
solve(n);
}
return 0;
}POJ3249 Test for Job 【DAG】+【记忆化搜索】,布布扣,bubuko.com
POJ3249 Test for Job 【DAG】+【记忆化搜索】
标签：poj3249
原文地址：http://blog.csdn.net/chang_mu/article/details/38664619
