数据结构OJ作业——最短路、拓扑排序

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poj3259 wormholes: http://poj.org/problem?id=3259

用spfa或者bellman-ford判断有无负环，我用的是spfa，使用cnt数组记录结点入队次数，次数大于等于n说明有负环。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAX = 6000;int n,m,w;struct T{    int to,w,nt;};T edge[MAX];int cnt[510],head[510],visit[510];int cnte;queue<int> q;void add(int u, int v , int w){    edge[cnte].to = v;    edge[cnte].w = w;    edge[cnte].nt = head[u];    head[u] = cnte;    cnte ++;}int spfa(void){    int dis[510];    memset(dis, 0x3f, sizeof(dis));    dis[1] = 0;    visit[1] = 1;    cnt[1] ++;    q.push(1);    while(!q.empty()) {        int now = q.front(); q.pop();        visit[now] = 0;        for (int i = head[now]; i; i = edge[i].nt) {            if (dis[now] + edge[i].w < dis[edge[i].to]) {                dis[edge[i].to] = dis[now] + edge[i].w;                if (!visit[edge[i].to]) {                    cnt[edge[i].to] ++;                    if (cnt[edge[i].to] >= n) return 1;                    q.push(edge[i].to);                    visit[edge[i].to] = 1;                }            }        }    }    return 0;}int main(int argc, char const *argv[]){    int f;    scanf("%d",&f);    while(f --) {        scanf("%d%d%d",&n,&m,&w);        cnte = 1;        memset(cnt,0,sizeof(cnt));        memset(visit,0,sizeof(visit));        memset(head,0,sizeof(head));        memset(edge,0,sizeof(edge));         int a,b,v;        for (int i = 0; i < m; i ++) {            scanf("%d%d%d",&a,&b,&v);            add(a,b,v);            add(b,a,v);        }        for (int i = 0; i < w; i ++) {            scanf("%d%d%d",&a,&b,&v);            add(a,b,-v);        }        if (spfa() ) {            puts("YES");        } else {            puts("NO");        }    }    return 0;}

poj1270 Following Orders: http://poj.org/problem?id=1270

求有向图的全部拓扑序列，这里用的是递归+回溯的方法，比较坑的是输出要求是按字典序排序，贡献两发WA

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <map>using  namespace std;const int MAX = 30;int degree[MAX], cnt, matr[MAX][MAX], vis[MAX];char s[MAX], ans[MAX];map<char, int> mp;void init(void){    cnt = 0;    mp.clear();    memset(matr, 0, sizeof(matr));    memset(vis, 0, sizeof(vis));    memset(s, 0, sizeof(s));    memset(degree, 0, sizeof(degree));}void dfstsort(int n); void solve(void){    dfstsort(1);}void dfstsort(int n){    if (n > cnt) {        for (int i = 1; i <= cnt; i ++) {            printf("%c",ans[i]);        }        printf("\n");        return ;    }    for (int i = 1; i <= cnt; i ++) {        if (!vis[i] && degree[i] == 0) {            ans[n] = s[i];            vis[i] = 1;            for (int j = 1; j <= cnt; j ++) {                if (matr[i][j]) {                    degree[j] --;                }            }            dfstsort(n + 1);            vis[i] = 0;            for (int j = 1; j <= cnt; j ++) {                if (matr[i][j]) {                    degree[j] ++;                }            }        }    }}int main(){    init();     char c, c1;    int isenter = 0;    int flag = 0, getc = 0;    while ((c = getchar()) != EOF) {        if (c == '\n') {            ++ flag;            // 字典序输出...             if (flag == 1) {                sort(s + 1, s + 1 + cnt);                for (int i = 1; i <= cnt; i ++) {                    mp[s[i]] = i;                }            }            if (flag == 2) {//              for (int i = 1; i <= cnt; i ++) {//                  cout << degree[i] << " " ;//              }                solve();                init();                flag = 0;                printf("\n");            }            continue;        }        if (c == ' ') continue;        if (flag == 0) {            s[++ cnt] = c;        }        if (flag == 1) {            if (getc == 0) {                c1 = c;                getc = 1;            } else {                matr[mp[c1]][mp[c]] = 1;                degree[mp[c]] ++;                getc = 0;            }        }    }    return 0;}

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