单源最短路（时间复杂度，传参数的注意）

 此题错在：传字符串参数会超时，找了好长时间的

Tom is playing a game called Idiomatic Phrases Game. An idiom consists of several Chinese characters and has a certain meaning. This game will give Tom two idioms. He should build a list of idioms and the list starts and ends with the two given idioms. For every two adjacent idioms, the last Chinese character of the former idiom should be the same as the first character of the latter one. For each time, Tom has a dictionary that he must pick idioms from and each idiom in the dictionary has a value indicates how long Tom will take to find the next proper idiom in the final list. Now you are asked to write a program to compute the shortest time Tom will take by giving you the idiom dictionary. 

Input

The input consists of several test cases. Each test case contains an idiom dictionary. The dictionary is started by an integer N (0 < N < 1000) in one line. The following is N lines. Each line contains an integer T (the time Tom will take to work out) and an idiom. One idiom consists of several Chinese characters (at least 3) and one Chinese character consists of four hex digit (i.e., 0 to 9 and A to F). Note that the first and last idioms in the dictionary are the source and target idioms in the game. The input ends up with a case that N = 0. Do not process this case. 

Output

One line for each case. Output an integer indicating the shortest time Tome will take. If the list can not be built, please output -1.

Sample Input

5
5 12345978ABCD2341
5 23415608ACBD3412
7 34125678AEFD4123
15 23415673ACC34123
4 41235673FBCD2156
2
20 12345678ABCD
30 DCBF5432167D
0

Sample Output

17
-1

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;

map<string,ll>m;
struct node{
	ll y;
	ll d;
	ll next_s;
}side[1000*1000];

ll cnt,head[1001],flag[1001],dis[1001],num[1001],cnt_s[1001];
vector<string>p;

bool judge(ll x,ll y,ll k){
	ll len=cnt_s[k];
	for(int i=len-4,j=0;j<4;i++,j++){
		if(p[x-1][i]!=p[y-1][j]){
			return false;
		}
	}
	 return true;
}

void init(){
	memset(head,-1,sizeof(head));
	cnt=0;
}

void add(ll x,ll y,ll d){
	side[cnt].y=y;
	side[cnt].d=d;
	side[cnt].next_s=head[x];
	head[x]=cnt++;
}

int main(){
	ll n;
	while(scanf("%lld",&n)&&n){
		for(int i=0;i<=n;i++){
			cnt_s[i]=0;
		}
		p.clear();
		init();
		ll t,cn=1;
		string s;
		for(int i=1;i<=n;i++){
			cin>>t>>s;
			cnt_s[cn]=s.size();
			num[cn++]=t;
			p.push_back(s);
		}
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++){
				//坑坑坑，直接传字符串复杂度会飙升，最好传字符床所在的下标
				if(judge(i,j,i)){ 
					add(i,j,num[i]);
				}
				if(judge(j,i,j)){
					add(j,i,num[j]);
				}
			}
		}
		for(int i=1;i<=n;i++){
			flag[i]=0;
			dis[i]=INF;
		}
		for(int i=head[1];i!=-1;i=side[i].next_s){
			dis[side[i].y]=side[i].d;
		}
		dis[1]=0;
		flag[1]=1;
		ll sign=0;
		for(int i=1;i<n;i++){
			ll min_s=INF,index;
			for(int j=1;j<=n;j++){
				if(!flag[j]&&dis[j]<min_s){
					min_s=dis[j];
					index=j;
				}
			}
			if(min_s==INF){
				sign=1;
				printf("-1\n");
				break;
			}
			if(index==n){
				break;
			}
			flag[index]=1;
			for(int k=head[index];k!=-1;k=side[k].next_s){
				if(!flag[side[k].y]&&dis[side[k].y]>dis[index]+side[k].d){
					dis[side[k].y]=dis[index]+side[k].d;
				}
			}
		}
		if(sign==0){
			printf("%lld\n",dis[n]);
		}
	}	
	return 0;
}