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POJ Big Christmas Tree(基础最短路)

                                  Big Christmas Tree


题目分析:

    叫你构造一颗圣诞树,使得 (sum of weights of all descendant nodes) × (unit price of the edge)尽量的小。转换后就是求根节点到每个节点的距离最短,也就是最短路。生成树可能会超时,我没试过。然后,求解最短路要用优化的解法不然会超时。最后的答案就是:sum = w[1] * dist[1] + w[2] * dist[2] + ..... w[n] * dist[n].可以自己推推样例就知道了。

   本来是一道简单的最短路。结果因为没有清空,浪费了一个早上的时间。,,,T_T

   以后,一定要记住啊!!!血的教训!!!!!


#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long LL;
const int MAXN = 50000 + 100;

struct Edge{
    int from,to;    
    LL c;
    Edge(){};
    Edge(int _f,int _t,LL _c)
        :from(_f),to(_t),c(_c){};
};

vector<Edge> edges;
vector<int> G[MAXN];
LL weight[MAXN],dist[MAXN];
bool vst[MAXN];
int numV,numE;

//清空
void init(){
    edges.clear();
    for(int i = 0;i <= numV;++i){
        G[i].clear();
    }
}
void addEdge(int x,int y,LL c){
    edges.push_back(Edge(x,y,c));
    int sz = edges.size();
    G[x].push_back(sz - 1);
}

//松弛操作
bool relax(int u,int v,LL c){
    if((dist[u] == -1) || (dist[u] > dist[v] + c)){
        dist[u] = dist[v] + c;
        return true;
    }
    return false;
}

//求解最短路
void spfa(LL src){
    int i,u,k;
    queue<int> Q;
    for(i = 0;i <= numV;++i){
        vst[i] = 0;
        dist[i] = -1;
    }

    Q.push(src);
    dist[src] = 0;

    while(!Q.empty()){
        u = Q.front();
        Q.pop();
        vst[u] = 0;
        for(i = 0;i < (int)G[u].size();++i){
            k = G[u][i];
            Edge& e = edges[k];
            if(relax(e.to,u,e.c) && !vst[e.to]){
                vst[e.to] = 1;
                Q.push(e.to);
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        int x,y;
        LL c;
        scanf("%d%d",&numV,&numE);
        init();
        for(int i = 1;i <= numV;++i){
            cin >> weight[i];
        }

        for(int i = 0;i < numE;++i){
            scanf("%d%d%I64d",&x,&y,&c);
            addEdge(x,y,c);
            addEdge(y,x,c);
        }

        spfa(1);
        LL sum = 0;
        bool flag = false;
        for(int i = 1;i <= numV;++i){
            if(dist[i] == -1){
                flag = true;
                break;
            }
            sum += dist[i] * weight[i];
        }
        if(flag)
            puts("No Answer");
        else
            printf("%I64d\n",sum);
    }
    return 0;
}




POJ Big Christmas Tree(基础最短路)