poj 1190 ROADS

poj 1190 ROADS

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12881 Accepted: 4756

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

#include 
#include 
#include 
#include 

using namespace std;

const int MAX_CITY = 100 + 5;
const int MAX_ROADS = 10000 + 5;
const int MAX_COST = 10000 + 5;
const int INF = 1 << 30;
// 总钱数，城市数，城市间路的条数
int K, N, R;
// 路线
struct Road {
    // 到达位置，长度和花费
    int D, L, T;
};
// 邻接表
vector<vector > cityMap(MAX_CITY);
// 最小路径长度
int minLen;
// 当前路线上的长度
int totalLen;
// 当前路线上的花费
int totalCost;
// 城市访问数组，0未访问，1访问过
int visited[MAX_CITY];
// minL[i][j]表示从城市1到当前城市i，花费为j的最短路径长度
int minL[MAX_CITY][MAX_COST];

void dfs(int S) {
    // 递归出口
    if(S == N) {
        minLen = min(minLen, totalLen);
        return;
    }

    for(int i = 0; i < cityMap[S].size(); i++) {
        // S到D有路
        int D = cityMap[S][i].D;
        if(!visited[D]) {
            // 走这条路到该城市的总花费
            int cost = totalCost + cityMap[S][i].T;
            // 最优性剪枝：当前花费已经大于拥有的钱了
            if(cost > K) {
                continue;
            }
            // 走这条路到该城市的总长度
            int len = totalLen + cityMap[S][i].L;
            // 最优性剪枝：当前长度已经大于等于最小路径长度或者
            // 大于等于从城市1到该城市，花费为cost的最短路径长度
            if(len >= minLen || len >= minL[D][cost]) {
                continue;
            }

            totalCost += cityMap[S][i].T;
            totalLen += cityMap[S][i].L;
            visited[D] = 1;
            minL[D][cost] = totalLen;
            dfs(D);
            // 回溯
            visited[D] = 0;
            totalCost -= cityMap[S][i].T;
            totalLen -= cityMap[S][i].L;
        }
    }
}

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;
        if(s != r.D) {
            cityMap[s].push_back(r);
        }
    }
    for(int i = 0; i < MAX_CITY; i++) {
        for(int j = 0; j < MAX_COST; j++) {
            minL[i][j] = INF;
        }
    }
    memset(visited, 0, sizeof(visited));
    totalCost = 0;
    totalLen = 0;
    minLen = INF;
    visited[1] = 1;
    dfs(1);
    if(minLen < INF) {
        cout << minLen << endl;
    } else {
        cout << -1 << endl;
    }

    return 0;
}