HDU - 4360As long as Binbin loves Sangsang最短路问题多状态记录

Binbin misses Sangsang so much. He wants to meet with Sangsang as soon as possible. 
Now Binbin downloads a map from ELGOOG.There are N (1<=N<=1,314) cities in the map and these cities are connected by M(0<=M<=13,520) bi-direct roads. Each road has a length L (1<=L<=1,314,520)and is marked using a unique ID, which is a letter fromthe string “LOVE”! 
Binbin rides a DONKEY, the donkey is so strange that it has to walk in the following sequence ‘L’->’O’->’V’->’E’->’L’->’O’->’V’->’E’->.... etc. 
Can you tell Binbin how far the donkey has to walk in order to meet with Sangsang? 
WARNING: Sangsang will feel unhappy if Binbin ride the donkey without a complete”LOVE” string. 
Binbin is at node 1 and Sangsang is at node N.

The first line has an integer T(1<=T<=520), indicate how many test cases bellow. 
Each test case begins with two integers N, M (N cities marked using an integer from 1…N and M roads). 
Then following M lines, each line has four variables“U V L letter”, means that there is a road between city U,V(1<=U,V<=N) with length L and the letter marked is‘L’,’O’,’V’ or ‘E’ 

For each test case, output a string 
1.  “Case ?: Binbin you disappoint Sangsang again, damn it!” 
If Binbin failed to meet with Sangsang or the donkey can’t finish a path withthe full “LOVE” string. 
2.  “Case ?: Cute Sangsang, Binbin will come with a donkey after travelling ? meters and finding ? LOVE strings at last.” 
Of cause, the travel distance should be as short as possible, and at the same time the “LOVE” string should be as long as possible. 

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!
        
        
/*
 *Author: 2486
 *Memory: 3376 KB		Time: 234 MS
 *Language: G++		Result: Accepted
 */

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int maxn = 2500;
const int maxm = 25000;
const LL INF = 100000000000000000;
int T, N, M, U, V, L;
char OP[5];
LL d[maxn][4], v[maxn][4];
bool vis[maxn];


typedef struct edge {
    int to;
    LL val;
    int OP;
    edge(int to,LL val,int OP):to(to), val(val), OP(OP) {}
} P;


vector<edge> G[maxm];


void init() {
    for(int i = 1; i <= N; i ++) G[i].clear();
}


void Dijkstra(int s) {
    queue<int> que;
    d[s][0] = 0;
    vis[s] = true;
    que.push(s);
    while(!que.empty()) {
        int s = que.front();
        que.pop();
        vis[s] = false;//消除记忆，可以向回走，只要有路
        for(int i = 0; i < G[s].size(); i ++) {
            edge e = G[s][i];
            int next = (e.OP + 1) % 4;//取下一个
            if(d[e.to][next] > d[s][e.OP] + e.val) {//代表着选择了当前这条路
                d[e.to][next] = d[s][e.OP] + e.val;
                v[e.to][next] = v[s][e.OP] + 1;
                if(vis[e.to]) continue;
                que.push(e.to);
                vis[e.to] = true;//标记暂时不能向前走
            } else if(d[e.to][next] == d[s][e.OP] + e.val) {
                v[e.to][next] = max(v[e.to][next], v[s][e.OP] + 1);
            }
        }
    }
}


int main() {
    //freopen("D://imput.txt","r",stdin);
    int tt = 0;
    scanf("%d",&T);
    while(T --) {
        tt ++;
        scanf("%d%d", &N, &M);
        init();
        int sv;
        int F[5] = {0};
        int cnt = 0;
        for(int i = 1; i <= N; i ++) {
            for(int j = 0; j < 4; j ++) {
                d[i][j] = INF;
                v[i][j] = 0;
            }
            vis[i] = false;
        }
        
        
        for(int i = 1; i <= M; i ++) {
            scanf("%d%d%d%s", &U, &V, &L, OP);
            if(OP[0] == 'L')sv = 0;
            else if(OP[0] == 'O')sv = 1;
            else if(OP[0] == 'V')sv = 2;
            else if(OP[0] == 'E')sv = 3;
            G[U].push_back( P(V, L, sv));
            G[V].push_back( P(U, L, sv));
            if(U == 1 && V == 1 && N == 1) {//起点终点一样，要特殊处理
                if(!F[sv]) {
                    F[sv] = L;
                    cnt ++;
                } else {
                    F[sv] = min(F[sv], L);
                }
            }
        }
        
        
        printf("Case %d: ", tt);
        if(cnt == 4) {//如果正好能够组成"LOVE"则输出
            LL sum = F[0] + F[1] + F[2] + F[3];
            printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding 1 LOVE strings at last.\n", sum);
            continue;
        }
        Dijkstra(1);
        if(v[N][0] < 4 || d[N][0] == INF) {//不满足条件
            printf("Binbin you disappoint Sangsang again, damn it!\n");
        } else {
            printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %I64d LOVE strings at last.\n",d[N][0], v[N][0] / 4);
        }
    }
    return 0;
}