Problem Description
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.
Input
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’
Output
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.
Sample Input
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
Sample Output
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!
題意：起點爲1，終點爲n，求起點到終點的最小距離，如果有相同的距離輸出最多的love個數，要求走的路徑必須嚴格按照L->O->V->E->L,且到達重點是必須是完整的LOVE串（即不存在類似LOVEL這種不是完整的LOVE的狀態）
將LOVE分別映射到0,1,2,3
首先發現每個點有不同的狀態，走L,走O,走V,走E，那麼一維的dis數組就不能完全將狀態記錄，所以->開個二維的dis數組記錄狀態，dis[i][j]表示i點走含有j的邊後距起點的最小距離
然後在spfa的過程中，根據邊的限制刷新最短路即可
#include<bits/stdc++.h>
#define INF 1000000000000000000
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int N=1500;
const int M=1314525;
ll n,m,maxx,minn,first[N],tot,num[N][4],vis[N][4];
ll dis[N][4];
struct node
{
ll v,w,op,next;
} g[M*2];
struct p
{
p() {}
p(ll a,ll b,ll c,ll d)
{
num=a,val=b,op=c,step=d;
}
ll num,val,op,step;
};
void add(ll u,ll v,ll w,ll op)
{
g[tot].v=v;
g[tot].w=w;
g[tot].op=op;
g[tot].next=first[u];
first[u]=tot++;
}
ll f(char op)
{
switch(op)
{
case 'L':
return 0;
case 'O':
return 1;
case 'V':
return 2;
case 'E':
return 3;
}
}
bool spfa()
{
queue<p>q;
for(int i=1; i<=n; i++)
for(int j=0; j<4; j++)dis[i][j]=INF;
mem(num,0);
q.push(p(1,0,3,0));
//初始時加入狀態爲1點，E邊，步數爲0，這樣可以在spfa選取的時候首先選擇爲L的邊
while(!q.empty())
{
p u=q.front();
q.pop();
for(ll i=first[u.num]; ~i; i=g[i].next)
{
if(g[i].op!=(u.op+1)%4)continue;
ll v=g[i].v;
//可以鬆弛時，將該點入隊
//當距離相同但可以增大當前選擇的字母數時，同樣也要將該點入隊，因爲此時後續由該點推出的num也會改變
if(dis[v][g[i].op]>u.val+g[i].w||dis[v][g[i].op]==u.val+g[i].w&&num[v][g[i].op]<u.step+1)
{
dis[v][g[i].op]=u.val+g[i].w;
num[v][g[i].op]=u.step+1;
q.push(p(v,dis[v][g[i].op],g[i].op,num[v][g[i].op]));
}
}
}
if(dis[n][3]==INF)return 0;
return 1;
}
int main()
{
ll t,x,y,w,q=0;
char op;
ios::sync_with_stdio(0);
cin>>t;
while(t--)
{
mem(first,-1);
tot=0;
cin>>n>>m;
for(ll i=0; i<m; i++)
{
cin>>x>>y>>w>>op;
add(x,y,w,f(op));
add(y,x,w,f(op));
}
printf("Case %I64d:",++q);
if(!spfa())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",dis[n][3],num[n][3]/4);
}
return 0;
}