BZOJ 2763: [JLOI2011]飞行路线 最短路

题目链接:

http://www.lydsy.com/JudgeOnline/problem.php?id=2763

题意:

题解:

最短路，在转移dis的时候多开一维k就好了
dis[i][j]->dis[t][j]+e[i][t]
dis[i][j]->dis[t][j+1]  if j<k

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

struct node{
    int x,y;
};

vector<node> E[maxn];
ll dis[maxn][15];
int inq[maxn][15];

int main(){
    int n,m,k,s,t;
    scanf("%d%d%d%d%d",&n,&m,&k,&s,&t);
    for(int i=0; i<m; i++){
        int u,v,w; scanf("%d%d%d",&u,&v,&w);
        E[u].push_back(node{v,w});
        E[v].push_back(node{u,w});
    }

    for(int i=0; i<=n; i++)
        for(int j=0; j<=k; j++)
            dis[i][j] = INF;
    memset(inq,0,sizeof(inq));
    queue<node> q;
    q.push((node){s,0});
    inq[s][0] = 1, dis[s][0] = 0;
    while(!q.empty()){
        node now = q.front();
        q.pop(); inq[now.x][now.y] = 0;
        for(int i=0; i<(int)E[now.x].size(); i++){
            int v = E[now.x][i].x, w = E[now.x][i].y;
            if(dis[v][now.y] > dis[now.x][now.y]+w){ // 不用票
                dis[v][now.y] = dis[now.x][now.y]+w;
                if(!inq[v][now.y]){
                    inq[v][now.y] = 1;
                    q.push((node){v,now.y});
                }
            }
            if(dis[v][now.y+1]>dis[now.x][now.y] && now.y<k){ // 用票
                dis[v][now.y+1] = dis[now.x][now.y];
                if(!inq[v][now.y+1]){
                    inq[v][now.y+1] = 1;
                    q.push((node){v,now.y+1});
                }
            }
        }
    }

    ll ans = INF;
    for(int i=0; i<=k; i++)
        if(dis[t][i] < ans)
            ans = dis[t][i];

    cout << ans << endl;

    return 0;
}