SPFA + 树形DP：The Ghost Blows Light

最新推荐文章于 2019-10-06 08:14:45 发布

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，同时着重介绍了AI音视频处理的应用场景和实现方法，如语义识别、语音变声等，为开发者提供全面的技术指南。

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1017 Accepted Submission(s): 303

Problem Description

My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road should cost some time). There is exactly one route between any two rooms, and each room contains some treasures. Now I am located at the 1st room and the exit is located at the Nth room.
Suddenly, alert occurred! The tomb will topple down in T minutes, and I should reach exit room in T minutes. Human beings die in pursuit of wealth, and birds die in pursuit of food! Although it is life-threatening time, I also want to get treasure out as much as possible. Now I wonder the maximum number of treasures I can take out in T minutes.

Input

There are multiple test cases.
The first line contains two integer N and T. (1 <= n <= 100, 0 <= T <= 500)
Each of the next N - 1 lines contains three integers a, b, and t indicating there is a road between a and b which costs t minutes. (1<=a<=n, 1<=b<=n, a!=b, 0 <= t <= 100)
The last line contains N integers, which Ai indicating the number of treasure in the ith room. (0 <= Ai <= 100)

Output

For each test case, output an integer indicating the maximum number of treasures I can take out in T minutes; if I cannot get out of the tomb, please output "Human beings die in pursuit of wealth, and birds die in pursuit of food!".

Sample Input

Sample Output

Source

2012 ACM/ICPC Asia Regional Changchun Online

题意：一张无向图，每条边走过去要花一定时间，每个节点上有一定数量的宝藏，在规定时间必须由起点走到终点，并尽可能多拿宝藏

分析：此题分两步做，首先求最短路，然后把这条路上的所有边的花费时间，再把总的时间减去上述时间
这样做的原因是最短路方法求出来的路是必经之路
然后做树形DP，以DFS的方法从起始点开始深搜，每搜到一个点x就更新一次所有dp[x][i]（指到x点时还剩i时间，这时最多能已经拿到多少宝藏，i从m到2倍到下个点的时间）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>

using namespace std;

int n, m;
const int INF = 1 << 31 - 1;

struct node{
    int to;
    int next;
    int weight;
}e[201];

int w[210];
int lastshow[210];
int dp[210][510]; //dp[x][i]表示从x出发时间还剩m的时候最多拿多少宝藏
bool inqueue[2010];
queue<int>q;
long long d[210];
int cnt = -1;
int used[210];
int path[210];

void insert(int a, int b, int w){
    e[++cnt].to = b;         //插入边，起始编号为0
    e[cnt].weight = w;
    e[cnt].next = lastshow[a];
    lastshow[a] = cnt;
}

bool spfa(){
    q.push(1);
    int qq[2005];
    while(!q.empty()){
        int x = q.front();
        q.pop();
        inqueue[x] = false;
        int id = lastshow[x];
        while(id != -1){
            if(d[x] < INF && d[e[id].to] > e[id].weight + d[x]){
                d[e[id].to] = e[id].weight + d[x];
                path[e[id].to] = x;           //记录一路上经过的点
                qq[e[id].to] = id;            //记录以上述点为终点的边
                if(!inqueue[e[id].to]){
                    inqueue[e[id].to] = true;
                    q.push(e[id].to);
                }
            }
            id = e[id].next;
        }
    }
    int i;
    for(i = n; path[i] != -1; i = path[i]){   //特别注意此处判断循环停止的条件是path[i]!=-1，若是i!=-1，会强制删除边0
        e[qq[i]].weight = 0;
    }
    return false;
}

void init(){
    int i;
    memset(used, 0, sizeof(used));
    memset(path, -1, sizeof(path));
    memset(dp, 0, sizeof(dp));
    memset(lastshow, -1, sizeof(lastshow));
    for(i = 1; i <= n; i ++){
        inqueue[i] = false;
    }
    for(i = 1; i <= n; ++ i)
        d[i]=INF;
    d[1]=0;
    cnt=-1;
    while(!q.empty()){
        q.pop();
    }
}

void dfs(int x){//树形dp
    used[x] = 1;
    for(int k = lastshow[x]; k!= -1; k = e[k].next){
        int y = e[k].to;
        if(used[y])
            continue;
        dfs(y); //每到一个新的点就深度遍历此点
        int tmp = e[k].weight * 2;
        for(int i = m; i >= tmp; i --){
            for(int j = i - tmp; j >= 0; j --){
                dp[x][i] = max(dp[x][i], dp[x][j] + dp[y][i - tmp - j]);
            }
        }
    }
    for(int i = 0; i <= m; i ++){
        dp[x][i] += w[x];//更新所有以x为起始的情况下能获得的宝藏数
    }
}

int main(){
    int i, j, k;
    while(~scanf("%d %d", &n, &m)){
        int a, b, c;
        init();
        for(i = 0; i < n - 1; i ++){
            scanf("%d %d %d", &a, &b, &c);
            insert(a, b, c);
            insert(b, a, c);
        }
        for(i = 1; i <= n; i ++){
            scanf("%d", &w[i]);
        }
        spfa();      //找出最短路，并把路上要花费的时间置为零，因为它是必经之路
        //cout << d[n] << endl;
        if(d[n] > m){
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
        }
        else{
            m -= d[n]; //更新还剩的时间
            dfs(1); 
            printf("%d\n", dp[1][m]);
        }
    }
    return 0;
}