数据结构OJ作业——最小生成树

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poj2485 Highways : http://poj.org/problem?id=2485

最小生成树输出最大边权

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 510;const int INF = 0x3f3f3f3f;int map[MAXN][MAXN];int n;int visit[MAXN];inline int Min(int a, int b){    return a > b? b : a;}int prim(void){    int ans = 0;    int lowcost[MAXN];    memset(visit,0,sizeof(visit));    for (int i = 2; i <= n; i ++) {        lowcost[i] = map[1][i];    }    lowcost[1] = 0;    visit[1] = 1;    for (int i = 1; i < n; i ++) {        int min = INF, minid;        for (int j = 1; j <= n ; j ++) {            if (!visit[j] && lowcost[j] < min) {                min = lowcost[j];                minid = j;            }        }        visit[minid] = 1;        if (ans < lowcost[minid]) {            ans = lowcost[minid];        }        for (int j = 1; j <= n ; j ++) {            lowcost[j] = Min(lowcost[j], map[minid][j]);        }    }    return ans;}int main(int argc, char const *argv[]){    int t;    scanf("%d",&t);    while (t --) {        scanf("%d",&n);        for (int i = 1; i <= n; i ++) {            for (int j = 1; j <= n; j ++) {                scanf("%d",&map[i][j]);            }        }        int ans = prim();        printf("%d\n",ans);    }    return 0;}

poj2377 Bad Cowtractors : http://poj.org/problem?id=2377

最大生成树，和最小生成树是一样的

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;int n, m, father[1005];struct edge {    int u,v,w;} e[20005];int find(int x){    while (x != father[x]) {        father[x] = father[father[x]];        x = father[x];    }    return x;}void merge(int x, int y){    x = find(x);    y = find(y);    if (x != y) {        father[x] = y;    }    return ;}bool cmp(const edge &e1, const edge &e2){    return e1.w > e2.w;}int main(){    while (~scanf("%d%d",&n,&m)) {        for (int i = 1; i <= n; i ++) {            father[i] = i;        }        int u, v, w;        for (int i = 1; i <= m; i ++) {            scanf("%d%d%d",&e[i].u, &e[i].v, &e[i].w);        }        sort(e + 1, e + 1 + m, cmp);        int cnt = 0, ans = 0;        for (int i = 1; i <= m; i ++) {            if (find(e[i].u) != find(e[i].v) ){                merge(e[i].u, e[i].v);                ans += e[i].w;                cnt ++;            }            if (cnt >= n - 1) break;         }        if (cnt >= n - 1)            printf("%d\n", ans);        else             puts("-1");    }    return 0;}

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