动态规划：HDU1224-Free DIY Tour

   Free DIY Tour

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6882    Accepted Submission(s): 2241

  

Problem Description

Weiwei is a software engineer of ShiningSoft. He has just excellently fulfilled a software project with his fellow workers. His boss is so satisfied with their job that he decide to provide them a free tour around the world. It's
a good chance to relax themselves. To most of them, it's the first time to go abroad so they decide to make a collective tour.

The tour company shows them a new kind of tour circuit - DIY circuit. Each circuit contains some cities which can be selected by tourists themselves. According to the company's statistic, each city has its own interesting point. For instance, Paris has its
interesting point of 90, New York has its interesting point of 70, ect. Not any two cities in the world have straight flight so the tour company provide a map to tell its tourists whether they can got a straight flight between any two cities on the map. In
order to fly back, the company has made it impossible to make a circle-flight on the half way, using the cities on the map. That is, they marked each city on the map with one number, a city with higher number has no straight flight to a city with lower number. 

Note: Weiwei always starts from Hangzhou(in this problem, we assume Hangzhou is always the first city and also the last city, so we mark Hangzhou both 1 andN+1),
and its interesting point is always 0.

Now as the leader of the team, Weiwei wants to make a tour as interesting as possible. If you were Weiwei, how did you DIY it?

 

Input

The input will contain several cases. The first line is an integer T which suggests the number of cases. Then T cases follows.

Each case will begin with an integer N(2 ≤ N ≤ 100) which is the number of cities on the map.

Then N integers follows, representing the interesting point list of the cities.

And then it is an integer M followed by M pairs of integers [Ai, Bi] (1 ≤ i ≤ M). Each pair of [Ai, Bi] indicates that a straight flight is available from City Ai to City Bi.

 

Output

For each case, your task is to output the maximal summation of interesting points Weiwei and his fellow workers can get through optimal DIYing and the optimal circuit. The format is as the sample. You may assume that there is only
one optimal circuit. 

Output a blank line between two cases.

 

Sample Input

2
3
0 70 90
4
1 2
1 3
2 4
3 4
3
0 90 70
4
1 2
1 3
2 4
3 4

 

Sample Output

CASE 1#
points : 90
circuit : 1->3->1

CASE 2#
points : 90
circuit : 1->2->1

解题心得：

1、这个题的大意很多点，每个点有一定的数值，部分点之间有边，要求从1点走到n+1点求最大的和。其实就是一个dp加了一些图论的问题，先按照起点拍一个序，然后标记一下终点是否可以走到，边走边标记，只有能够从1点走出达到的点才能加上。题意还要求记录一下路径，可以将dp数组开成一个结构体，在状态转移的时候顺便记录一下转移前的点的位置。最后再从n+1点倒过来找一下就可以了（实现方法看代码）。

2、还有一种做法就是将这个问题弄成纯dp，看成求最大上升子序列的和。

dp+图论代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 110;
struct DP
{
    int sum;
    int pre_city;
}dp[maxn];
struct
{
    bool flag;
    int Num;
}
num[maxn];
struct node
{
    int start;
    int End;
}maps[maxn*10000];

bool cmp(node a,node b)
{
    return a.start < b.start;
}

//这个函数就是拿来找路径的
void get_path(int n)
{
    printf("circuit : ");
    stack <int> s;
    int End,pre;
    End = n;
    while(dp[End].pre_city != 0)//层层返回，从终点找路径找到起点
    {
        s.push(End);
        End = dp[End].pre_city;
    }
    s.push(1);//将起点手动压入

    printf("%d",s.top());
    s.pop();
    while(!s.empty())
    {
        if(s.top() == n)//到达的其实是n+1点，但是n+1点其实就是起点，这里处理一下
        {
            s.pop();
            s.push(1);
        }
        printf("->");
        printf("%d",s.top());
        s.pop();
    }
    printf("\n");
}

int main()
{
    int t;
    scanf("%d",&t);
    int Test = 1;
    while(t--)
    {
        int n,m,T;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i].Num);
            num[i].flag = false;
        }
        num[1].flag = true;
        scanf("%d",&m);
        T = 0;
        while(m--)
        {
            scanf("%d%d",&maps[T].start,&maps[T].End);
            if(maps[T].start >= maps[T].End)//起点比终点更大，不计入
                T--;
            T++;
        }

        sort(maps,maps+T,cmp);//按照起点排一个序
        for(int i=0;i<T;i++)
        {
            if(num[maps[i].start].flag)//要能从1走到该点才行
                num[maps[i].End].flag = true;
            if(dp[maps[i].End].sum < dp[maps[i].start].sum + num[maps[i].End].Num && num[maps[i].start].flag)
            {
                dp[maps[i].End].sum = dp[maps[i].start].sum + num[maps[i].End].Num;
                dp[maps[i].End].pre_city = maps[i].start;//记录一下转移过来的路径
            }
        }
        printf("CASE %d#\n",Test++);
        printf("points : %d\n",dp[n+1]);

        get_path(n+1);

        //每次消除上次计算的，不然WA很惨
        if(t != 0)
            printf("\n");
        for(int i=0;i<=T;i++)
            maps[i].End = maps[i].start = 0;
        for(int i=0;i<=n+1;i++)
        {
            dp[i].pre_city = 0;
            dp[i].sum = 0;
            num[i].flag = false;
            num[i].Num = 0;
        }
    }
}

纯dp代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 110;
bool maps[maxn][maxn];
int dp[maxn];
int num[maxn],pre[maxn];

void get_path(int n)
{
    printf("circuit : ");
    stack <int> s;
    int N = n;

    while(dp[N] != 0)//倒过来找到路径
    {
        s.push(N);
        N = pre[N];
    }
    s.push(1);
    printf("%d",s.top());
    s.pop();
    while(!s.empty())//再正过来输出
    {
        printf("->");
        if(s.top() == n)
        {
            s.pop();
            s.push(1);
        }
        printf("%d",s.top());
        s.pop();
    }
    printf("\n");
}

int main()
{
    int T = 1;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        memset(maps,0,sizeof(maps));
        memset(num,0,sizeof(num));
        memset(pre,0,sizeof(pre));
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        int m;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            if(a < b)
                maps[a][b] = true;
        }

        for(int i=1;i<=n+1;i++)
            for(int j=1;j<i;j++)
            {
                if(maps[j][i])//有这条路
                {
                    if(dp[i] < dp[j] + num[i])//状态转移
                    {
                        dp[i] = dp[j] + num[i];
                        pre[i] = j;//记录转移的位置
                    }
                }
            }

        printf("CASE %d#\n",T++);
        printf("points : %d\n",dp[n+1]);
        get_path(n+1);
        if(t != 0)
            printf("\n");
    }
}