题意
Farmer John已经决定把水灌到他的n(1<=n<=300)块农田，农田被数字1到n标记。把一块土地进行灌水有两种方法，从其他农田饮水，或者这块土地建造水库。 建造一个水库需要花费Wi(1<=Wi<=100000),连接两块土地需要花费Pij(1<=pij<=100000,pij=pji,pii=0). 计算Farmer John所需的最少代价。
思路
很经典的最小生成树模型……很久没做最小生成树一下子没想出来TAT……
首先得有个水源吧，设个虚节点当水源，然后连向每个土地，权值为建造水库的花费，其他的照常建图，然后求最小生成树就行了。
代码
[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, end) for (int i = begin; i <= end; i ++)
using namespace std;
const int MAX = 305;
struct edge{
int v, w;
edge(){
}
edge(int _v, int _w){
v = _v;
w = _w;
}
};
struct MST{
vector <edge> adj[MAX];
int dist[MAX];
bool vis[MAX];
priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > PQ;
void init(int n){
for (int i = 0; i <= n; i ++){
adj[i].clear();
vis[i] = false;
}
}
void add_edge(int u, int v, int w){
adj[u].push_back(edge(v, w));
adj[v].push_back(edge(u, w));
}
int solve(int s, int n){
for (int i = 0; i <= n; i ++) dist[i] = 0x3fffffff;
while(!PQ.empty()) PQ.pop();
dist[s] = 0;
PQ.push(make_pair(0, s));
while(!PQ.empty()){
int u = PQ.top().second;
PQ.pop();
vis[u] = true;
for (int i = 0; i < (int)adj[u].size(); i ++){
int v = adj[u][i].v, w = adj[u][i].w;
if (!vis[v] && dist[v] > w){
dist[v] = w;
PQ.push(make_pair(w, v));
}
}
}
int res = 0;
for (int i = 1; i <= n; i ++) res += dist[i];
return res;
}
}prim;
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n;
scanf("%d", &n);
prim.init(n+1);
for (int i = 1; i <= n; i ++){
int tmp;
scanf("%d", &tmp);
prim.add_edge(n+1, i, tmp);
}
for (int i = 1; i <= n; i ++){
for (int j = 1; j <= n; j ++){
int tmp;
scanf("%d", &tmp);
if (i >= j) continue;
prim.add_edge(i, j, tmp);
}
}
printf("%d\n", prim.solve(n+1, n+1));
return 0;
}
[/cpp]
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