如果是指标准模板库(stl)中容器的一般pushback()操作函数，那么是指在容器尾端插入一项数据，比如
vector<int> a(10);
a.pushback(10);
那么a的尾端，同时也是唯一一个数据a[0]就会为设置为10。
题目：
Description
N cities named with numbers 1 … N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins).
Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash.
We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has.
Input
The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way.
The second line contains the integer N, 2 <= N <= 100, the total number of cities.
The third line contains the integer R, 1 <= R <= 10000, the total number of roads.
Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters :
S is the source city, 1 <= S <= N
D is the destination city, 1 <= D <= N
L is the road length, 1 <= L <= 100
T is the toll (expressed in the number of coins), 0 <= T <=100
Notice that different roads may have the same source and destination cities.
Output
The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins.
If such path does not exist, only number -1 should be written to the output.
Sample Input
5
6
7
1 2 2 3
2 4 3 3
3 4 2 4
1 3 4 1
4 6 2 1
3 5 2 0
5 4 3 2
Sample Output
11
Source
CEOI 1998
解答：
/*
//POJ1724
//ROADS 深度优先
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int K,N,R;//N个城市
struct Road{
int d,L,t;
};
//用邻接表存放整条路的信息
vector< vector<Road> >G(110);
int minLen;//存储目前找过的最佳的路径长度
int totalLen;//已经探索的路径已经走了多长
int totalCost;//长在走的路已经花了多少钱
int visited[110];//标志数组，表示一个城市是否已经走过
void dfs(int s){
if(s==N){
minLen=min(minLen,totalLen);
return;
}
for(int i=0;i<G[s].size();++i){
Road r=G[s][i];
if(totalCost+r.t>K)
continue;
if(!visited[r.d]){
totalLen+=r.L;
totalCost+=r.t;
visited[r.d]=1;
dfs(r.d);
visited[r.d]=0;
totalLen-=r.L;
totalCost-=r.t;
}
}
}
int main(){
cin>>K>>N>>R;
for(int i=0;i<R;i++){
int s;
Road r;
cin>>s>>r.d>>r.L>>r.t;//起点s，终点d，边的长的L，边的过路费t
if(s!=r.d){
G[s].push_back(r);
}
}
//初始化
memset(visited,0,sizeof(visited));
totalLen=0;
minLen=1<<30;
totalCost=0;
visited[1]=1;
dfs(1);
if(minLen<(1<<30)){
cout<<minLen<<endl;
}else
cout<<"-1"<<endl;
return 0;
} //超时
*/
/*
//POJ1724
//ROADS 深度优先
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int K,N,R;//N个城市
struct Road{
int d,L,t;
};
//用邻接表存放整条路的信息
vector< vector<Road> >G(110);
int minLen;//存储目前找过的最佳的路径长度
int totalLen;//已经探索的路径已经走了多长
int totalCost;//长在走的路已经花了多少钱
int visited[110];//标志数组，表示一个城市是否已经走过
void dfs(int s){
if(s==N){
minLen=min(minLen,totalLen);
return;
}
for(int i=0;i<G[s].size();++i){
Road r=G[s][i];
if(totalCost+r.t>K)
continue;
if(!visited[r.d]){
if(totalLen+r.L>=minLen){//////////最优性剪枝：
continue;//////////////////////（1）如果当前已经找到的最优路径长度为L，那么再继续搜索的过程中，总长度已经大于等于L的走法，就可以直接放弃，不用坚持到底
} /////////////////////////////////很强的剪枝，效果拔群
totalLen+=r.L;
totalCost+=r.t;
visited[r.d]=1;
dfs(r.d);
visited[r.d]=0;
totalLen-=r.L;
totalCost-=r.t;
}
}
}
int main(){
cin>>K>>N>>R;
for(int i=0;i<R;i++){
int s;
Road r;
cin>>s>>r.d>>r.L>>r.t;//起点s，终点d，边的长的L，边的过路费t
if(s!=r.d){
G[s].push_back(r);
}
}
//初始化
memset(visited,0,sizeof(visited));
totalLen=0;
minLen=1<<30;
totalCost=0;
visited[1]=1;
dfs(1);
if(minLen<(1<<30)){
cout<<minLen<<endl;
}else
cout<<"-1"<<endl;
return 0;
} //可惜还是超时
*/
//POJ1724
//ROADS 深度优先
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int K,N,R;//N个城市
struct Road{
int d,L,t;
};
//用邻接表存放整条路的信息
vector< vector<Road> >G(110);
int minL[110][10010];///////////////////////////////////
int minLen;//存储目前找过的最佳的路径长度
int totalLen;//已经探索的路径已经走了多长
int totalCost;//长在走的路已经花了多少钱
int visited[110];//标志数组，表示一个城市是否已经走过
//用mid[k][m]表示：走到城市k时总过路费为m的条件下，最优路径的长度。若在后续的搜索中，在此走到k时，如果总路费恰好为m，且此时的路径长度已经超过mid[k][m]，则不必再走下去了。
void dfs(int s){
if(s==N){
minLen=min(minLen,totalLen);
return;
}
for(int i=0;i<G[s].size();++i){
Road r=G[s][i];
if(totalCost+r.t>K)
continue;
if(!visited[r.d]){
if(totalLen+r.L>=minLen){
continue;
}
/////////////////////////////////////////////////////////////可行性剪枝：
if(totalLen+r.L>=minL[r.d][totalCost+r.t])///////////////////啥都不说了：
continue;/////////////////////////////////////////////////加速超快！！
minL[r.d][totalCost+r.t]=totalLen+r.L;////////////////////////
totalLen+=r.L;
totalCost+=r.t;
visited[r.d]=1;
dfs(r.d);
visited[r.d]=0;
totalLen-=r.L;
totalCost-=r.t;
}
}
}
int main(){
cin>>K>>N>>R;
for(int i=0;i<R;i++){
int s;
Road r;
cin>>s>>r.d>>r.L>>r.t;//起点s，终点d，边的长的L，边的过路费t
if(s!=r.d){
G[s].push_back(r);
}
}
//初始化
memset(visited,0,sizeof(visited));
totalLen=0;
minLen=1<<30;
totalCost=0;
visited[1]=1;//////////////////////////////////
for(int i=0;i<110;i++){/////////////////////////
for(int j=0;j<10010;j++){/////////////////
minL[i][j]=1<<30;///////////////////////
}//////////////////////////////////////////
}////////////////////////////////////////////////
dfs(1);
if(minLen<(1<<30)){
cout<<minLen<<endl;
}else
cout<<"-1"<<endl;
return 0;
}