HDU 4360 As long as Binbin loves Sangsang(最短路)

题意:问从1到n的最短路,所走的路必须是LOVELOVE,而且是完整的LOVE。就是道最短路的题。

这题比较坑,让人无语。

提供测试数据吧;

2
1 0
1 4
1 1 9 L
1 1 5 O
1 1 2 V
1 1 3 E
答案:

Binbin you disappoint Sangsang again, damn it!

Cute Sangsang, Binbin will come with a donkey after travelling 19 meters and finding 1 LOVE strings at last.

还有,这里的有一组数据超int ,要用long long。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;
const int N = 1320;
const int M = 13529;
const int INF =0x3f3f3f3f;
int n,m;
long long dis[N][5];
bool visit[N][5];
struct LL
{
    int to,dis,se,nex;
} L[M<<1];
int F[N],cnt;
void add(int f,int t,int d,int se)
{
    L[cnt].dis = d;
    L[cnt].se = se;
    L[cnt].to = t;
    L[cnt].nex = F[f];
    F[f] = cnt++;
}
void init()
{
    scanf("%d%d",&n,&m);
    int f,t,d,se;
    memset(F,0,sizeof(F));
    cnt= 1;
    char tt[3];
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%d%s",&f,&t,&d,tt);
        if(tt[0]=='L') se = 0;
        if(tt[0]=='O') se = 1;
        if(tt[0]=='V') se = 2;
        if(tt[0]=='E') se = 3;
        add(f,t,d,se);
        add(t,f,d,se);
    }
}
struct nod
{
    int v,se,num;long long dis;
    bool operator<(const nod t) const
    {
        if(dis==t.dis) return num<t.num;
        return dis>t.dis;
    }
};
priority_queue<nod> que;
int nums[N][5];
void solve(int c)
{
    while(!que.empty()) que.pop();
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<5;j++) dis[i][j] = 0x3f3f3f3f3f3f3f3fll;
    }
    memset(visit,false,sizeof(visit));
    memset(nums,0,sizeof(nums));
    nod e,t;
    e.v=1,e.dis=0,e.se=0,e.num=0;
    que.push(e);
    dis[1][0] = INF;
    while(!que.empty())
    {
        e = que.top();
        que.pop();
        if(visit[e.v][e.se]) continue;
        visit[e.v][e.se] = true;
        if(e.dis==0)
        {
            visit[e.v][e.se] = false;
        }
        //dis[e.v][e.se] = e.dis;
        if(e.v==n&&e.se ==0&&e.dis>0)
        {
            cout<<"Case "<<c<<": Cute Sangsang, Binbin will come with a donkey after travelling "<<e.dis<<" meters and finding "<<e.num<<" LOVE strings at last."<<endl;
            //printf("Case %d: Cute Sangsang, Binbin will come with a donkey after travelling %d meters and finding %d LOVE strings at last.\n",c++,e.dis,e.num);
            return ;
        }
        for(int i=F[e.v]; i; i=L[i].nex)
        {
            int to = L[i].to;
            if(L[i].se!=e.se) continue;
            int se= (e.se+1)%4;
            if(se==0) t.num = e.num+1;
                else t.num = e.num;
            if(!visit[to][se]&&(dis[to][se]>e.dis+L[i].dis||(nums[to][se]<t.num&&dis[to][se]==e.dis+L[i].dis)))
            {
                dis[to][se]=e.dis+L[i].dis;
                t.dis = dis[to][se];
                t.se = se;
                t.v = to;

                nums[to][se] = t.num;
                que.push(t);
            }
        }
    }
    //printf("Case %d: Binbin you disappoint Sangsang again, damn it!\n",c);
    cout<<"Case "<<c<<": Binbin you disappoint Sangsang again, damn it!"<<endl;
}
int main()
{
    freopen("in.txt","r",stdin);
    int cas,T=1;
    scanf("%d",&cas);
    while(cas--)
    {
        init();
        solve(T++);
    }
    return 0;
}


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