As long as Binbin loves Sangsang

2
4 4
1 2 1 L
2 1 1 O
1 3 1 V
3 4 1 E
4 4
1 2 1 L
2 3 1 O
3 4 1 V
4 1 1 E

Case 1: Cute Sangsang, Binbin will come with a donkey after travelling 4 meters and finding 1 LOVE strings at last.
Case 2: Binbin you disappoint Sangsang again, damn it!

题意描述：

n个点，m条边，起点为1，终点为n（注意！n可能等于1）的无向图（注意！无向图）。只能按照L->O->V->E->L->O->...的顺序走，求到达n点的最短长度和组成LOVE的最大个数。


解题思路：

SPFA,多一个dp数组记录，特判n=1的情况，路径长度可能超int。


参考代码：

#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const double eps=1e-10;
const long long INF=1e17;
const int MAXN=1500;
typedef long long LL;
struct edge
{
    int to,love;
    LL value;
    edge(int a,LL b,int c)
    {
        to=a;
        value=b;
        love=c;
    }
};

vector<edge>G[MAXN];
int n,m,dp[MAXN][4];
LL mincost[MAXN][4];
bool used[MAXN];

void SPFA(int start)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<4;j++)
        {
            mincost[i][j]=INF;
            dp[i][j]=0;
        }
        used[i]=false;
    }
    queue<int>q;
    q.push(start);
    mincost[start][0]=0;
    used[start]=true;
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        used[now]=false;
        for(int i=0;i<G[now].size();i++)
        {
            edge e=G[now][i];
            int next=(e.love+1)%4;
            if(mincost[e.to][next]>mincost[now][e.love]+e.value)
            {
                mincost[e.to][next]=mincost[now][e.love]+e.value;
                dp[e.to][next]=dp[now][e.love]+1;
                if(!used[e.to])
                {
                    q.push(e.to);
                    used[e.to]=true;
                }
            }
            else if(mincost[e.to][next]==mincost[now][e.love]+e.value)
                dp[e.to][next]=max(dp[e.to][next],dp[now][e.love]+1);
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int tcase,f=0;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int spe=0;
        LL spec[4];
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            G[i].clear();
        memset(spec,0,sizeof(spec));
        for(int i=1;i<=m;i++)
        {
            int from,to,lov;
            LL val;
            char l[4];
            scanf("%d%d%lld%s",&from,&to,&val,l);
            if(l[0]=='L')
                lov=0;
            else if(l[0]=='O')
                lov=1;
            else if(l[0]=='V')
                lov=2;
            else if(l[0]=='E')
                lov=3;
            G[from].push_back(edge(to,val,lov));
            G[to].push_back(edge(from,val,lov));
            if(n==1)//特判只有一个点的情况
            {
                if(!spec[lov])
                {
                    spec[lov]=val;
                    spe++;
                }
                else
                    spec[lov]=min(spec[lov],val);
            }
        }
        printf("Case %d: ",++f);
        if(n==1)
        {
            if(spe==4)
            {
                LL sum=spec[0]+spec[1]+spec[2]+spec[3];
                printf("Cute Sangsang, Binbin will come with a donkey after travelling %lld meters and finding 1 LOVE strings at last.\n",sum);
            }
            else
                printf("Binbin you disappoint Sangsang again, damn it!\n");
            continue;
        }
        SPFA(1);
        if(mincost[n][0]==INF||dp[n][0]<4)//只需要计算完整的LOVE
            printf("Binbin you disappoint Sangsang again, damn it!\n");
        else
            printf("Cute Sangsang, Binbin will come with a donkey after travelling %lld meters and finding %d LOVE strings at last.\n",mincost[n][0],dp[n][0]/4);
    }
    return 0;
}

HDU 4360 As long as Binbin loves Sangsang（SPFA）

原文：http://blog.csdn.net/noooooorth/article/details/47319163