As long as Binbin loves SangsangTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4094 Accepted Submission(s): 880 Problem Description Binbin misses Sangsang so much. He wants to meet with Sangsang as soon as possible. Input The first line has an integer T(1<=T<=520), indicate how many test cases bellow. Output For each test case, output a string 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! Author FZU Source 2012 Multi-University Training Contest 7 Recommend zhuyuanchen520 | We have carefully selected several similar problems for you: 2426 3790 3435 3917 4309 |
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int MAXN = 2500;
const int MAXM = 25000;
#define INF 0x3f3f3f3f3f3f3f3f //long long 就定义更大的
long long dis[MAXN][4], cnt[MAXN][4];
bool vis[MAXN];
queue<int>q;
struct Node
{
int to;
long long w;
int letter;
};
vector<Node> G[MAXM];
void addedge(int u,int v,long long w,int letter)
{
Node t={v,w,letter};
G[u].push_back(t);
}
void init(int n)
{
for(int i = 1; i <= n; i ++) G[i].clear();
}
void spfa(int s,int n)//单源最短路(s为起点,n为节点总数)
{
int u;
for(int i=0;i<=n;i++)
for(int j=0;j<4;j++)
{
dis[i][j]=INF;
cnt[i][j]=0;
}
memset(vis,0,sizeof(vis));
while(!q.empty())
q.pop();
q.push(s);
vis[s]=true;
dis[s][0]=0;
while(!q.empty())
{
u=q.front();
q.pop();
vis[u]=false;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i].to;
long long w=G[u][i].w;
int letter=G[u][i].letter;
int next_letter=(letter+1)%4;
if(dis[v][next_letter]>dis[u][letter]+w) //原来的一个点变成四个点
{ //同样是唯一确定的之前状态
dis[v][next_letter]=dis[u][letter]+w;
cnt[v][next_letter]=cnt[u][letter]+1; //累积一定是按LOVE顺序累积个数的,所以cnt[u][0]/4就能表示几个LOVE
if(!vis[v]) //cnt是从0开始的,所以9个数,只是8
{ //cnt[u][0]前面必须是E才有值,不然就是0.
vis[v]=true;
q.push(v);
}
}
else if(dis[v][next_letter]==dis[u][letter]+w)
{
cnt[v][next_letter]=max(cnt[v][next_letter],cnt[u][letter]+1);
}
}
}
}
int main()
{
int T;
int n,m;
int u,v;
long long w;
int Case = 1;
char str;
scanf("%d",&T);
while(T--)
{
scanf("%d%d", &n, &m);
init(n);
int letter;
long long F[5] = {0};
int Count = 0;
for(int i = 1; i <= n; i ++)
{
for(int j = 0; j < 4; j ++)
{
dis[i][j] = INF;
cnt[i][j] = 0;
}
vis[i] = false;
}
for(int i = 1; i <= m; i ++)
{
scanf("%d %d %lld %c", &u, &v, &w,&str);
if(str == 'L')
letter=0;
else if(str == 'O')
letter=1;
else if(str == 'V')
letter=2;
else if(str == 'E')
letter=3;
addedge(u,v,w,letter);
addedge(v,u,w,letter);
if(u== 1 && v == 1 && n == 1) {//起点终点一样,要特殊处理
if(!F[letter])
{
F[letter] = w;
Count++;
} else
{
F[letter] = min(F[letter], w);
}
}
}
printf("Case %d: ", Case++);
if(Count == 4)
{//如果正好能够组成"LOVE"则输出
long long 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;
}
spfa(1,n);
if(cnt[n][0]==0||dis[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",dis[n][0], cnt[n][0]/4);
}
}
return 0;
}