题干：
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 题目大意：
题意第一行给了一个k,代表你有k的钱数，下一行有一个n,代表n个点，然后一个m，代表m条边，然后接下来m行,每行有四个数，分别代表起点、终点、路径长度和要花费的钱数，题目想问在花的钱不超过k的情况下，从1---n的最短路径是多少。
解题报告：
注意cost和length不能直接存到二维数组中，然后用min来去重边，，因为可能一个大一个小，emmm对于这种两个权值的还是要注意啊。。。
AC代码1：（94ms）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 500 + 5;
const int INF = 0x3f3f3f3f;
int k,n,m;
struct Node {
int v;
int w,c;
int ne;
Node(){}
Node(int v,int w,int c):v(v),w(w),c(c){}
}e[MAX*MAX];
int head[MAX];
bool vis[MAX];
int ans = INF,tot;
void add(int u,int v,int w,int c) {
e[++tot].v = v;
e[tot].w = w;
e[tot].c = c;
e[tot].ne = head[u];
head[u] = tot;
}
void dfs(int cur,int c,int w) {//c花费 w路程
if(c > k || w > ans) return;
if(cur == n) {
ans = min(ans,w);
return ;
}
for(int i = head[cur]; i!=-1; i=e[i].ne) {
Node v = e[i];
if(vis[v.v]) continue;
vis[v.v] = 1;
dfs(v.v,c+v.c,w+v.w);
vis[v.v] = 0;
}
}
int main()
{
cin>>k>>n>>m;
int u,v,w,c;
memset(head,-1,sizeof head);
for(int i = 1; i<=m; i++) {
scanf("%d%d%d%d",&u,&v,&w,&c);
add(u,v,w,c);
}
dfs(1,0,0);
if(ans == INF) puts("-1");
else printf("%d\n",ans);
return 0 ;
} AC代码2：（32ms）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 500 + 5;
const int INF = 0x3f3f3f3f;
int k,n,m;
struct Node {
int v;
int dis,c;
int ne;
Node() {}
Node(int v,int w,int c):v(v),dis(dis),c(c) {}
} e[MAX*MAX];
struct Point {
int id,dis,cost;
bool friend operator < (Point a,Point b) {
return a.dis>b.dis;
}
};
//vector<Node> vv[MAX];
int head[MAX];
bool vis[MAX];
int ans = INF,tot;
void add(int u,int v,int w,int c) {
e[++tot].v = v;
e[tot].dis = w;
e[tot].c = c;
e[tot].ne = head[u];
head[u] = tot;
}
ll dp[105][10005];//dp[i][j] : 到i号城市，花费为j，的最短路。
void Dijkstra() {
for(int i=1; i<=n; i++)
for(int j=0; j<=k; j++)
dp[i][j]=INF;
dp[1][0]=0;
priority_queue<Point>pq;
Point now,cur;
cur.id=1;
cur.cost=0;
cur.dis=0;
pq.push(cur);
while(!pq.empty()) {
cur=pq.top();
pq.pop();
if(cur.id==n) {
printf("%d\n",cur.dis);
return;
}
for(int i=head[cur.id]; i!=-1; i=e[i].ne) {
int cost=cur.cost+e[i].c;
if(cost>k)continue;
if(dp[e[i].v][cost]>cur.dis+e[i].dis) {
dp[e[i].v][cost]=cur.dis+e[i].dis;
now.id=e[i].v;
now.cost=cost;
now.dis=dp[e[i].v][cost];
pq.push(now);
}
}
}
puts("-1");
}
int main() {
cin>>k>>n>>m;
int u,v,w,c;
memset(head,-1,sizeof head);
for(int i = 1; i<=m; i++) {
scanf("%d%d%d%d",&u,&v,&w,&c);
add(u,v,w,c);
}
Dijkstra();
return 0 ;
} 写Dijkstra的时候就不需要if(vis[cur.id]) continue;了。。（其实这一句在一般的Dijkstra中也可以不写，只是统计路径条数的时候必须要写上这个，不然就算重复了（比如那道qduoj的题））
AC代码3：（47 - 32ms）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 500 + 5;
const int INF = 0x3f3f3f3f;
int k,n,m;
struct Node {
int v;
int dis,c;
int ne;
Node() {}
Node(int v,int w,int c):v(v),dis(dis),c(c) {}
} e[MAX*MAX];
struct Point {
int id,dis,cost;
bool friend operator < (Point a,Point b) {
return a.dis>b.dis;
}
};
//vector<Node> vv[MAX];
int head[MAX];
bool vis[MAX];
int ans = INF,tot;
void add(int u,int v,int w,int c) {
e[++tot].v = v;
e[tot].dis = w;
e[tot].c = c;
e[tot].ne = head[u];
head[u] = tot;
}
ll dp[105][10005];//dp[i][j] : 到i号城市，花费为j，的最短路。
int bfs() {
priority_queue<Point>pq;
Point now,cur;
cur.id=1;
cur.cost=0;
cur.dis=0;
pq.push(cur);
while(!pq.empty()) {
cur=pq.top(); pq.pop();
if(cur.id==n) {
if(cur.cost <= k) return cur.dis;
}
for(int i=head[cur.id]; i!=-1; i=e[i].ne) {
int nowcost=cur.cost+e[i].c;
if(nowcost>k)continue;
now.id=e[i].v;
now.cost=nowcost;
now.dis=cur.dis + e[i].dis;
pq.push(now);
}
}
puts("-1");
}
int main() {
cin>>k>>n>>m;
int u,v,w,c;
memset(head,-1,sizeof head);
for(int i = 1; i<=m; i++) {
scanf("%d%d%d%d",&u,&v,&w,&c);
add(u,v,w,c);
}
printf("%d\n",bfs());
return 0 ;
} 总结：
对于poj我还能说什么，，换上vector立刻tle。。
AC代码4：（A*算法，0ms）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 150 + 5;
const int INF = 0x3f3f3f3f;
int k,n,m;
struct Node {
int v;
int dis,c;
int ne;
Node() {}
Node(int v,int w,int c):v(v),dis(dis),c(c) {}
} e[MAX*MAX],r[MAX*MAX];
struct Point {
int id,dis,cost;
bool friend operator < (Point a,Point b) {
return a.dis>b.dis;
}
};
//vector<Node> vv[MAX];
int head[MAX],read[MAX];
bool vis[MAX];
int Dis[MAX];
int ans = INF,tote,totr;
void add(int u,int v,int w,int c) {
e[++tote].v = v;
e[tote].dis = w;
e[tote].c = c;
e[tote].ne = head[u];
head[u] = tote;
}
void add2(int u,int v,int w,int c) {
r[++totr].v = v;
r[totr].dis = w;
r[totr].c = c;
r[totr].ne = read[u];
read[u] = totr;
}
void Dijkstra() {
memset(Dis,INF,sizeof Dis);
priority_queue<Point>pq;
Point now,cur;
cur.id=n;cur.cost=0;cur.dis=0;
Dis[n] = 0;
pq.push(cur);
while(pq.size()) {
cur=pq.top(); pq.pop();
if(vis[cur.id]) continue;
vis[cur.id] = 1;
for(int i=read[cur.id]; i!=-1; i=r[i].ne) {
int cost=cur.cost+r[i].c;
// if(cost>k)continue;因为就是求标准的最短路，所以这个就没啥用了先
if(Dis[r[i].v]>cur.dis+r[i].dis) { //松弛条件
Dis[r[i].v]=cur.dis+r[i].dis;
now.id=r[i].v;now.cost=cost;now.dis=Dis[r[i].v];
pq.push(now);
}
}
}
}
struct A {
int id,dis,c;
A(){};
A(int id,int dis,int c):id(id),dis(dis),c(c){}
bool friend operator < (A a,A b) {
return a.dis + Dis[a.id] > b.dis + Dis[b.id];
}
};
int Astar(int st,int ed) {
priority_queue< A > pq;
pq.push(A(st,0,0));
while(pq.size()) {
A cur = pq.top();pq.pop();
if(cur.id == ed) return cur.dis;
for(int i = head[cur.id]; i != -1; i = e[i].ne) {
if(cur.c + e[i].c <= k) pq.push(A(e[i].v,cur.dis + e[i].dis,cur.c + e[i].c));
}
}
return -1;
}
int main() {
cin>>k>>n>>m;
int u,v,w,c;
memset(head,-1,sizeof head);
memset(read,-1,sizeof read);
for(int i = 1; i<=m; i++) {
scanf("%d%d%d%d",&u,&v,&w,&c);
add(u,v,w,c);
add2(v,u,w,c);
}
Dijkstra();
printf("%d\n",Astar(1,n));
return 0 ;
} 总结：
其实这个代码求最短路的那个Point结构体都不需要带上cost这个参数，，，因为没啥用。
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