标签：normal memset 最大权值 sort ada clu nes oss ++i
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 10137 | Accepted: 2348 |
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
题目链接:点击打开链接
给出n个点的权值和m条路径, 问一条路的最大权值是多少.
起点入度为0, 对全部入度为0的点赋值到dp数组, 进行拓扑排序, 最后遍历出度为0的点, 求得最大值.
AC代码:
#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
int n, m, num;
int cost[MAXN], in[MAXN], out[MAXN], head[MAXN], dp[MAXN];
bool vis[MAXN];
struct node
{
/* data */
int fr, to, nxt;
}e[MAXN * 10];
void add(int x, int y)
{
e[num].fr = x;
e[num].to = y;
e[num].nxt = head[x];
head[x] = num++;
}
void toposort()
{
int cnt = 1;
while(cnt < n) {
for(int i = 1; i <= n; ++i)
if(in[i] == 0 && !vis[i]) {
vis[i] = true;
cnt++;
for(int j = head[i]; j != -1; j = e[j].nxt) {
int x = e[j].to;
in[x]--;
if(dp[i] + cost[x] > dp[x]) dp[x] = dp[i] + cost[x];
}
}
}
}
int main(int argc, char const *argv[])
{
while(scanf("%d%d", &n, &m) != EOF) {
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(head, -1, sizeof(head));
memset(vis, false, sizeof(vis));
num = 1;
for(int i = 1; i <= n; ++i)
scanf("%d", &cost[i]);
for(int i = 1; i <= m; ++i) {
int x, y;
scanf("%d%d", &x, &y);
add(x, y);
in[y]++;
out[x]++;
}
for(int i = 1; i <= n; ++i)
if(in[i] == 0) dp[i] = cost[i];
else dp[i] = -INF;
toposort();
int ans = -INF;
for(int i = 1; i <= n; ++i)
if(out[i] == 0 && dp[i] > ans) ans = dp[i];
printf("%d\n", ans);
}
return 0;
}标签：normal memset 最大权值 sort ada clu nes oss ++i
原文地址：http://www.cnblogs.com/liguangsunls/p/6912251.html
