hdu 2544

最短路

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 50493    Accepted Submission(s): 22211

Problem Description

在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？

 

Input

输入包括多组数据。每组数据第一行是两个整数N、M（N<=100，M<=10000），N表示成都的大街上有几个路口，标号为1的路口是商店所在地，标号为N的路口是赛场所在地，M则表示在成都有几条路。N=M=0表示输入结束。接下来M行，每行包括3个整数A，B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A与路口B之间有一条路，我们的工作人员需要C分钟的时间走过这条路。
输入保证至少存在1条商店到赛场的路线。

 

Output

对于每组输入，输出一行，表示工作人员从商店走到赛场的最短时间

 

Sample Input

2 1
1 2 3
3 3
1 2 5
2 3 5
3 1 2
0 0

 

Sample Output

3
2

 

从简单的题目开始敲起

#include
#include
#include
#include
#include

using namespace std;

typedef long long ll;
const ll inf = 1e16;
const int maxn = 200 + 5;
int n, m;
struct Edge{
    int from, to;
    int val;
    Edge(int from, int to, int val): from(from), to(to), val(val){}
};
vector  edges;
vector  G[maxn];
struct node{
    int u;
    ll d;
    node(int u, ll d): u(u), d(d){}
    bool operator < (const node &a) const{
        return d > a.d;
    }
};
ll d[maxn];
bool vis[maxn];

void init(){
    edges.erase(edges.begin(), edges.end());
    for (int i = 0; i <= n; i++){
        G[i].erase(G[i].begin(), G[i].end());
    }
    for (int i = 1; i <= m; i++){
        int u, v, val;
        scanf("%d%d%d", &u, &v, &val);
        edges.push_back(Edge(u, v, val));
        edges.push_back(Edge(v, u, val));
        int l = edges.size();
        G[u].push_back(l - 2);
        G[v].push_back(l - 1);
    }
}

void dijkstra(){
    for (int i = 0; i <= n; i++){
        vis[i] = false;
        d[i] = inf;
    }
    priority_queue  que;
    d[1] = 0;
    que.push(node(1, d[1]));
    while (!que.empty()){
        node x = que.top();
        que.pop();
        int u = x.u;
        if (vis[u] == true) continue;
        vis[u] = true;
        for (int i = 0; i < G[u].size(); i++){
            Edge y = edges[G[u][i]];
            if (d[y.to] > d[u] + y.val){
                d[y.to] = d[u] + y.val;
                que.push(node(y.to, d[y.to]));
            }
        }
    }
    printf("%I64d\n", d[n]);
}

int main(){
    while (scanf("%d%d", &n, &m) && n + m != 0){
        init();
        dijkstra();
    }
    return 0;
}







#include
#include
#include
#include
#include

using namespace std;

typedef long long ll;
const ll inf = 1e16;
const int maxn = 200 + 5;
int n, m;
ll d[maxn][maxn];

int main(){
    while (scanf("%d%d", &n, &m) && n + m != 0){
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= n; j++){
                d[i][j] = inf;
            }
        }
        for (int i = 0; i < m; i++){
            int u, v, val;
            scanf("%d%d%d", &u, &v, &val);
            if (val < d[u][v] && val < d[v][u])
                d[u][v] = d[v][u] = val;
        }
        for (int i = 1; i <= n; i++){
            d[i][i] = 0;
        }
        for (int k = 1; k <= n; k++){
            for (int i = 1; i <= n; i++){
                for (int j = 1; j <= n; j++){
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                }
            }
        }
        printf("%I64d\n", d[1][n]);
    }
    return 0;
}






#include
#include
#include
#include
#include

using namespace std;

typedef long long ll;
const ll inf = 1e16;
const int maxn = 200 + 5;
int n, m;
struct Edge{
    int from, to, val;
    Edge(int from, int to, int val): from(from), to(to), val(val){}
};
vector  edges;
vector  G[maxn];
int cnt[maxn];
int vis[maxn];
ll d[maxn];

void init(){
    edges.erase(edges.begin(), edges.end());
    for (int i = 1; i <= n; i++){
        G[i].erase(G[i].begin(), G[i].end());
        cnt[i] = 0;
        vis[i] = false;
        d[i] = inf;
    }
    for (int i = 0; i < m; i++){
        int u, v, d;
        scanf("%d%d%d", &u, &v, &d);
        edges.push_back(Edge(u, v, d));
        edges.push_back(Edge(v, u, d));
        int l = edges.size();
        G[u].push_back(l - 2);
        G[v].push_back(l - 1);
    }
}

bool bellman(int s){
    queue  que;
    que.push(s);
    d[s] = 0;
    vis[s] = true;
    while (!que.empty()){
        int u = que.front();
        que.pop();
        vis[u] = false;
        for (int i = 0; i < G[u].size(); i++){
            Edge &y = edges[G[u][i]];
            if (d[y.to] > d[u] + y.val){
                d[y.to] = d[u] + y.val;
                if (!vis[y.to]){
                    que.push(y.to);
                    vis[y.to] = true;
                    if (++cnt[y.to] == n) return false;
                }
            }
        }

    }
    return true;
}

int main(){
    while (scanf("%d%d", &n, &m) && n + m != 0){
        init();
        bool flag = bellman(1);
        if (flag){
            printf("%I64d\n", d[n]);
        }
    }
    return 0;
}