HNACM(八)D-引水工程

来源：互联网发布：微信公众号对接数据库编辑：程序博客网时间：2024/04/27 19:50

传送门

最小生成树算法

首先找到一个自建水库最少费用的然后以此为根本找到与之相关的边
还要注意当权值相等时的处理方法

#include <bits/stdc++.h> #define N 310#define ll long long#define MAX 11111using namespace std;int n, w[N], p[N][N], water[N];//w[N]是 每个地方自建水库所需的费用//water[N]是存放已经解决用水问题的地方 bool vis[N];int prim(int pos, int sum){    int i, j, k, cur = 0;    memset(vis, false, sizeof(vis));    memset(water, 0, sizeof(water));    water[cur++] = pos;    vis[pos] = true;    while(cur < n){        int mindist = MAX;        int min1 = MAX;        int u = -1;        for (i = 0; i < cur; i++){//遍历所有的已经解决用水问题的点            for (j = 0; j < n; j++){                if (!vis[j] && j != water[i]){                    if (mindist > p[water[i]][j]){ //未解决问的点，找出从其他地方引水的最小费用                         mindist = p[water[i]][j];                        min1 = w[j];                        u = j;                    }else if (mindist == p[water[i]][j] && min1 > w[j]){//如果当前最小值和权值相等，比较自建费用                         min1 = w[j];                        u = j;                    }                }            }        }        vis[u] = true;        water[cur++] = u;        sum += min(min1, mindist);//去自建费用和从其他地方引水的费用的较小者     }    return sum;}int main(){#ifndef ONLINE_JUDGE//  freopen("1.txt", "r", stdin);//  freopen("d.in", "r", stdin);//  freopen("d1.out", "w", stdout);#endif      int i, j, k, K, Min, pos;    scanf("%d", &K);    while(K--){        scanf("%d", &n);        Min = MAX;        for (i = 0; i < n; i++){            scanf("%d", &w[i]);            if (Min > w[i]){//先寻找自建水库的费用最小的地方                 pos = i;                Min = w[i];            }        }        for (i = 0; i < n; i++){            for (j = 0; j < n; j++){                scanf("%d", &p[i][j]);            }        }        printf("%d\n", prim(pos, Min));//先从自建水库最小的地方开始     }    return 0;}

另一种方法，把自建费用当成结点0到结点i的费用，然后用最下生成树算法

感谢@Dart 提供思路

#include <iostream>#include <cstdio>#define MAXN 310using namespace std;int g[MAXN][MAXN], dis[MAXN];bool vis[MAXN];const int INF = 0x3f3f3f3f;int n;int prim(int st){    int i, j, k;    for (i = 0; i <= n; i++){        vis[i] = false;        dis[i] = g[st][i];    }    dis[st] = 0;    vis[st] = true;    int sum = 0;    for (i = 1; i <= n; i++){        int Min = INF;        int v = -1;        for (j = 0; j <= n; j++){            if (!vis[j] && Min > dis[j]){                v = j;                Min = dis[j];            }        }        vis[v] = true;        sum += Min;        for (j = 0; j <= n; j++){            if (!vis[j] && g[v][j] < dis[j]){                dis[j] = g[v][j];            }        }    }    return sum;}int main(){    int i, j, k, T;    scanf("%d", &T);    while(T--){        scanf("%d", &n);        for (i = 1; i <= n; i++){            scanf("%d", &g[0][i]);            g[i][0] = g[0][i];        }        for (i = 1; i <= n; i++){            for (j = 1; j <= n; j++){                scanf("%d", &g[i][j]);            }        }        printf("%d\n", prim(0));    }    return 0;}

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