USACO 2009 Feb Gold 3.Revamping Trails

Description

Farmer John dutifully checks on the cows every day. He traverses so me of the M (1 <= M <= 50,000) trails conveniently numbered 1..M from pasture 1 all the way out to pasture N (a journey which is always possible for trail maps given in the test data). The N (1 <= N <= 10,000) pastures conveniently numbered 1..N on Farmer John’s farm are currently connected by bidirectional dirt trails. Each trail i connects pastures P1_i and P2_i (1 <= P1_i <= N; 1 <= P2_i <= N) and requires T_i (1 <= T_i <= 1,000,000) units of time to traverse.
He wants to revamp some of the trails on his farm to save time on his long journey. Specifically, he will choose K (1 <= K <= 20) trails to turn into highways, which will effectively reduce the trail’s traversal time to 0. Help FJ decide which trails to revamp to minimize the resulting time of getting from pasture 1 to N.

Input

• Line 1: Three space-separated integers: N, M, and K

• Lines 2..M+1: Line i+1 describes trail i with three space-separated
  integers: P1_i, P2_i, and T_i

Output

• Line 1: The length of the shortest path after revamping no more than
  K edges

Sample Input

4 4 11 2 102 4 101 3 13 4 100

Sample Output

1

Key To Problem

小小的图论+二维dpdis[u][num]表示到点u，翻新num条道路所需要走的最短的时间dis[to][num]=dis[u][num]+s;dis[to][num+1]=dis[u][num];需要用dijkstra+堆优化，据说SPFA超时，并没有尝试。

CODE

#include<queue>#include<vector>#include<cstdio>#include<cstring>#include<algorithm>#define N 10010using namespace std;vector<int>V[N];vector<int>pre[N];struct node{    int u,num,s;    bool friend operator <(node x,node y)    {        return x.s>y.s;    }};int n,m,l;int dis[N][30];bool used[N][30];void dijkstra(int u){    memset(dis,0x3f,sizeof(dis));    priority_queue<node>Q;    node k;    k.u=u,k.s=0,k.num=0;    Q.push(k);    dis[u][0]=0;    while(Q.size())    {        k=Q.top();        Q.pop();        if(used[k.u][k.num])            continue;        used[k.u][k.num]=1;        int num=k.num;        for(int i=0;i<V[k.u].size();i++)        {            int to=V[k.u][i];            int s=pre[k.u][i];            if(dis[to][num]>dis[k.u][num]+s)            {                dis[to][num]=dis[k.u][num]+s;                node p;                p.u=to,p.num=num,p.s=dis[to][num];                Q.push(p);            }            if(num+1<=l&&dis[to][num+1]>dis[k.u][num])            {                dis[to][num+1]=dis[k.u][num];                node p;                p.u=to,p.num=num+1,p.s=dis[to][num+1];                Q.push(p);            }        }    }}int main(){//  freopen("revamp.in","r",stdin);//  freopen("revamp.out","w",stdout);    scanf("%d%d%d",&n,&m,&l);    for(int i=1;i<=m;i++)    {        int x,y,z;        scanf("%d%d%d",&x,&y,&z);        V[x].push_back(y);        pre[x].push_back(z);        V[y].push_back(x);        pre[y].push_back(z);    }    dijkstra(1);    printf("%d\n",dis[n][l]);    return 0;}