2012 Multi-University Training Contest 7-1001 hdu4360 As long as Binbin loves Sangsang

每条边除了有边权以外，还有一个字母标记。标记可以是“LOVE”里面任意字符。

每个点，要拆成四个点，分别代表到达该点的标记为L,O,V,E的最短路。

第一步就是求最短路，直接Dijkstra就可以了。

trick在于，至少要找到一个LOVE串，在只有一个节点的时候，有几条自环，至少必须走LOVE四条自环。此时，必须另外加一个节点表示开始节点。

还有一个trick就是距离可能超过int。

1 2 1314520 L

1 2 1314520 O

1 2 1314520 V

2 3 1314520 E

3 4 1314520 L

3 4 1314520 O

3 4 1314520 V

4 5 1314520 E

...

这种情况下1313个点，2624条边，每条边长度1314520，并且每条边都必须走，所以，超int了（至少signed不够）。

这道题主要就是trick多，比赛的时候wrong answer了好多次，我基本把所有的trick都经历了一遍

下面是我ac的代码

#include <iostream>#include <cstdio>#include <algorithm>#include <queue>#include <list>using namespace std;typedef long long ll;const int M = 1500;const int N = 15000;ll INF = 1 << 30;struct node{int to;int w;int ch;bool bo;node *next;}d[M];node pool[2 * N];int n,m,s,t,top;ll dist[M][4];int num[M][4];void INIT();void Dijkstra();int change(char ch){if(ch == 'L')return 0;else if(ch=='O')return 1;else if(ch == 'V')return 2;else if(ch == 'E')return 3;}int main(){int ttt,cas=1;INF = INF * INF;scanf("%d",&ttt);while(ttt--){scanf("%d%d",&n,&m);INIT();Dijkstra();if(dist[n][0] == INF)printf("Case %d: Binbin you disappoint Sangsang again, damn it!\n",cas++);else{printf("Case %d: Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.\n",cas++,dist[t][0],num[t][0]);}}return 0;}void INIT(){ll u,v,w;char str[3];for(int i = 1;i <= M;i++){d[i].next = NULL;dist[i][0] = dist[i][1] = dist[i][2] = dist[i][3] = INF;}memset(num,0,sizeof(num));s = 1;t = n;top = 0;for(int i = 0;i < m;i++){scanf("%d%d%d",&u,&v,&w);scanf("%s",&str);pool[top].to = v;pool[top].w = w;pool[top].ch = change(str[0]);pool[top].next = d[u].next;d[u].next = &pool[top++];pool[top].to = u;pool[top].w = w;pool[top].ch = change(str[0]);pool[top].next = d[v].next;d[v].next = &pool[top++];}return;}void Dijkstra(){int flag,g,numlove;node *st;priority_queue< pair<ll,pair<int,int> > > qu;while(!qu.empty())qu.pop();qu.push(make_pair(0,make_pair(s,0)));while(!qu.empty()){flag = qu.top().second.first;g = qu.top().second.second;ll di = -qu.top().first;qu.pop();numlove = num[flag][g];st = d[flag].next;int y = (g +1) % 4;if(g + 1 == 4) numlove++;while(st!=NULL){long long x = st->ch;if(x == g){if(dist[st->to][y] > di + st->w){ dist[st->to][y] = di + st->w;num[st->to][y] = numlove;qu.push(make_pair(-dist[st->to][y],make_pair(st->to,y)));}else if(dist[st->to][y] == di + st->w&&numlove > num[st->to][y]){dist[st->to][y] = di + st->w;num[st->to][y] = max(numlove,num[st->to][y]);qu.push(make_pair(-dist[st->to][y],make_pair(st->to,y)));}}st = st->next;}}return;}