POJ1724 ROADS 解题报告【最短路/SPFA】

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
解题报告
这道题的题意是：n个点、m条边、初始k个点数，每条边有自己的点数、长度，问在可用点数的范围内走出的最短路，如果走不拢就输出-1。
我们将原先的dis数组变成二维，dis[u][k]代表走到第u个点，花费k个点数的最短路长度。我们知道，dis只有一维的时候他是这样更新的：dis[v]=min(dis[u]+ed[i].w)。那么现在改变了定义的dis数组又怎样更新呢？其实是这样的：dis[v][j]=min(dis[u][j-ed[i].t]+ed[i].w) , ed[i].t<=j<=k。
那么解法就很明了了。
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=100,M=10000,inf=2e9;
struct edge
{
    int v,l,t,next;
}ed[M+5];
int k,n,m;
int head[N+5],num;
int dis[N+5][M+5],flag[N+5];
void build(int u,int v,int l,int t)
{
    ed[++num].v=v;
    ed[num].l=l;
    ed[num].t=t;
    ed[num].next=head[u];
    head[u]=num;
}
void SPFA()
{
    queue<int>q;
    memset(flag,0,sizeof(flag));
    for(int i=1;i<=n;i++)
    for(int j=0;j<=k;j++)dis[i][j]=inf;
    for(int i=0;i<=k;i++)dis[1][i]=0;
    q.push(1);
    flag[1]=1;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        flag[u]=0;
        for(int i=head[u];i!=-1;i=ed[i].next)
        {
            int v=ed[i].v;
            for(int j=ed[i].t;j<=k;j++)
            {
                if(dis[v][j]>dis[u][j-ed[i].t]+ed[i].l)
                {
                    dis[v][j]=dis[u][j-ed[i].t]+ed[i].l;
                    if(!flag[v])
                    {
                        q.push(v);
                        flag[v]=1;
                    }
                }
            }
        }
    }
    int ans=inf;
    for(int i=0;i<=k;i++)
    ans=min(ans,dis[n][i]);
    if(ans==inf)printf("-1\n");
    else printf("%d\n",ans);
}
int main()
{
    memset(head,-1,sizeof(head));
    scanf("%d%d%d",&k,&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v,l,t;
        scanf("%d%d%d%d",&u,&v,&l,&t);
        build(u,v,l,t);
    }
    SPFA();
    return 0;
}