HDU 4360 As long as Binbin loves Sangsang

这题调到后面真是调疯了.....一直wa啊wa................卧槽, 尼玛的原来是手敲队列的时候, 队列大小开小了, 因为spfa一个结点能多次进入队列......这他妈都能错........卧槽...........

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
inline int Rint() { int x; scanf("%d", &x); return x; }
inline int max(int x, int y) { return (x>y)? x: y; }
inline int min(int x, int y) { return (x<y)? x: y; }
#define FOR(i, a, b) for(int i=(a); i<=(b); i++)
#define FORD(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(x) for(int i=0; i<(x); i++)
typedef __int64 int64;
//#define INF (1<<30)
const int64 INF = 0x7f7f7f7f7f7f7f7f;
const double eps = 1e-8;
#define bug(s) cout<<#s<<"="<<s<<" "

#define MAXN (1316*4)
#define MAXM (13522*2)
struct node {  int u, v, w; }a[MAXM];
int idx;
int head[MAXN], next[MAXM];
void init() { idx=0; memset(head, -1, sizeof(head)); }
void addedge(int u, int v, int w) { a[idx].u=u; a[idx].v=v; a[idx].w=w; next[idx]=head[u]; head[u]=idx++; }
int n, m;

int q[MAXN*10];		//	不是MAXN!!!因为一个点可以多次进入队列!!!............WA到死
int front, tail;
int inq[MAXN];	//在队中就不用再加了.		//这其实是bfs的变种

int64 step[MAXN];	//维护步数
int64 dist[MAXN];		// st->i

void spfa(int st)
{
	memset(step, 0, sizeof(step));
	memset(inq, 0, sizeof(inq));
	FOR(i, 1, n)								//1-th
		dist[i] = i==st? 0: INF;
	front = tail = 0;
	q[tail++] = st;
	inq[st] = 1;
	while(front<tail)
	{
		int u = q[front++];		//别忘了 front++, pop
		inq[u] = 0;
		for(int e=head[u]; e!=-1; e=next[e])
		{
			int v = a[e].v;
			if(dist[u]<INF && ( dist[v]>dist[u]+a[e].w || (dist[v]==dist[u]+a[e].w && step[u]+1>step[v]) ) )	//重点: 在最短路相等时, 选边数最多的!!!
			{
				step[v] = step[u]+1;
				dist[v] = dist[u]+a[e].w;
				//bug(v);bug(dist[v])<<endl;
				if(!inq[v])
				{
					q[tail++] = v;
					inq[v] = 1;
				}
			}
		}

	}

}

int main()
{
	int T = Rint();
	FOR(ca, 1,  T)
	{
		printf("Case %d: ", ca);
		init();	//清图
		n = Rint();
		m =Rint();
		REP(m)
		{
			int u = Rint(), v=Rint(), w=Rint();		//1-th
			char id[2];	scanf("%s",  id);

			//	LOVE  i, i+n, i+2n, i+3n.
			switch(id[0])
			{
			case 'L':
				addedge(u+3*n, v, w);
				addedge(v+3*n, u, w);
				break;
			case 'O':
				addedge(u, v+n, w);
				addedge(v, u+n, w);
				break;
			case 'V':
				addedge(u+n, v+2*n, w);
				addedge(v+n, u+2*n, w);
				break;
			case 'E':
				addedge(u+2*n, v+3*n, w);
				addedge(v+2*n, u+3*n, w);
				break;
			}
		}

		if(n == 1)		//特判
		{
			int64 ans = 0;
			int find = 1;
			FOR(u, 1, 4)	//枚举4个点
			{
				int64 minx = INF;
				for(int e=head[u]; e!=-1; e=next[e])	//枚举邻边
				{
					if(minx>a[e].w) minx = a[e].w;
				}
				if(minx == INF) { find=0; break; }
				ans+=minx;
			}
			if(find)
			{
				printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.\n", ans, 1);
			}
			else
				printf("Binbin you disappoint Sangsang again, damn it!\n");
			continue;
		}

		int tn = n;
		n*=4;
		m = idx;
		spfa(1+3*tn);
		if(dist[tn*4]<INF)		//题目要求必须走完整的 LOVE.
			printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %I64d LOVE strings at last.\n", dist[tn*4], step[tn*4]/4);
		else
			printf("Binbin you disappoint Sangsang again, damn it!\n");
	}
}