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]
