hdu 1102 uva 10397（最小生成树prim）

hdu 1102:

题意：

给一个邻接矩阵，给一些村庄间已经修的路，问最小生成树。

解析：

把已经修的路的权值改为0，套个prim（）。

注意prim 最外层循坏为n-1。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int maxn = 100 + 10;const int inf = 0x3f3f3f3f;const double eps = 1e-8;const double pi = 4 * atan(1.0);const double ee = exp(1.0);int n;int g[maxn][maxn];int prim(){    bool vis[maxn];    int dis[maxn];    memset(vis, false, sizeof(vis));    memset(dis, inf, sizeof(dis));    vis[1] = true;    dis[1] = 0;    int res = 0;    int mark = 1;    for (int i = 1; i <= n - 1; i++)///mark = 1 so n - 1    {        for (int j = 1; j <= n; j++)        {            if (!vis[j] && g[mark][j] < dis[j])                dis[j] = g[mark][j];        }        int mindis = inf;        for (int j = 1; j <= n; j++)        {            if (!vis[j] && dis[j] < mindis)            {                mindis = dis[j];                mark = j;            }        }        vis[mark] = true;        res += mindis;    }    return res;}int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);    #endif // LOCAL    while (~scanf("%d", &n))    {        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= n; j++)            {                scanf("%d", &g[i][j]);            }        }        int q;        scanf("%d", &q);        while (q--)        {            int u, v;            scanf("%d%d", &u, &v);            g[u][v] = g[v][u] = 0;        }        printf("%d\n", prim());    }    return 0;}

uva 10397:

基本上是一样的题目。。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int maxn = 750 + 10;const int inf = sqrt(2 * 20000 * 20000) + 10;int n;double g[maxn][maxn];double x[maxn], y[maxn];double dist(double x1, double y1, double x2, double y2){    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}double prim(){    double dis[maxn];    bool vis[maxn];    for (int i = 0; i <= n; i++)    {        dis[i] = inf;        vis[i] = false;    }    dis[1] = 0.0;    vis[1] = true;    double res = 0.0;    int mark = 1;    for (int i = 1; i <= n - 1; i++)    {        for (int j = 1; j <= n; j++)        {            if (!vis[j] && g[mark][j] < dis[j])                dis[j] = g[mark][j];        }        double mindis = inf;        for (int j = 1; j <= n; j++)        {            if (!vis[j] && dis[j] < mindis)            {                mindis = dis[j];                mark = j;            }        }        vis[mark] = true;        res += mindis;    }    return res;}int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);    #endif // LOCAL    while (~scanf("%d", &n))    {        for (int i = 1; i <= n; i++)        {            scanf("%lf%lf", &x[i], &y[i]);        }        for (int i = 1; i <= n; i++)        {            for (int j = i + 1; j <= n; j++)            {                g[i][j] = g[j][i] = dist(x[i], y[i], x[j], y[j]);            }        }        int q;        scanf("%d", &q);        for (int i = 0; i < q; i++)        {            int u, v;            scanf("%d%d", &u, &v);            g[u][v] = g[v][u] = 0;        }//        for (int i = 1; i <= n; i++)//        {//            for (int j = 1; j <= n; j++)//            {//                printf("%.2lf ", g[i][j]);//            }//            printf("\n");//        }        printf("%.2lf\n", prim());    }    return 0;}