HDU 4966 最小树形图

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <sstream>
 12 #include <iostream>
 13 #include <algorithm>
 14 #include <bitset>
 15 #include <climits>
 16 using namespace std;
 17 
 18 #define wh while
 19 #define inf (int)(~0u/2)
 20 #define FOR(i, n) for(int i = 0; i < n; i++)
 21 #define FOR1(i, n) for(int i = 1; i < n; i++)
 22 #define FOR2(i, n) for(int i = 0; i <= n; i++)
 23 #define REP(i,n) for(int i = 1; i <= n; i++)
 24 #define FORI(it,n) for(typeof(n.begin()) it = n.begin(); it != n.end(); it++)
 25 #define sf scanf
 26 #define pf printf
 27 #define frs first
 28 #define sec second
 29 #define psh push_back
 30 #define mkp make_pair
 31 #define PB(x) push_back(x)
 32 #define MP(x, y) make_pair(x, y)
 33 #define clr(abc,z) memset(abc,z,sizeof(abc))
 34 #define lt(v) v << 1
 35 #define rt(v) v << 1 | 1
 36 //#define mid ((l + r) >> 1)
 37 //#define lson l, mid, v << 1
 38 //#define rson mid + 1, r, v << 1 | 1
 39 
 40 #define fre freopen("1.txt", "r", stdin)
 41 #define freout freopen("1.txt", "w", stdout)
 42 
 43 typedef long long LL;
 44 typedef long double LD;
 45 const int maxn = 600 + 5;
 46 const int INF = 1e9;
 47 // 固定根的最小树型图，邻接矩阵写法，时间复杂度O（n^3）
 48 struct MDST {
 49     int n;
 50     int w[maxn][maxn]; // 边权
 51     int vis[maxn];     // 访问标记，仅用来判断无解
 52     int ans;           // 计算答案
 53     int removed[maxn]; // 每个点是否被删除
 54     int cid[maxn];     // 所在圈编号
 55     int pre[maxn];     // 最小入边的起点
 56     int iw[maxn];      // 最小入边的权值
 57     int max_cid;       // 最大圈编号
 58 
 59     void init(int n) {
 60         this->n = n;
 61         for(int i = 0; i < n; i++)
 62             for(int j = 0; j < n; j++) w[i][j] = INF;
 63     }
 64 
 65     void AddEdge(int u, int v, int cost) {
 66         w[u][v] = min(w[u][v], cost);
 67     }
 68 
 69     int dfs(int s) {
 70         vis[s] = 1; int ans = 1;
 71         for(int i = 0; i < n; i++)
 72             if(!vis[i] && w[s][i] < INF) ans += dfs(i);
 73         return ans;
 74     }
 75     bool cycle(int u) {
 76         max_cid++; int v = u;
 77         while(cid[v] != max_cid) {
 78             cid[v] = max_cid; v = pre[v];
 79         }
 80         return v == u;
 81     }
 82     void update(int u) {
 83         iw[u] = INF;
 84         for(int i = 0; i < n; i++)
 85             if(!removed[i] && w[i][u] < iw[u]) {
 86                 iw[u] = w[i][u]; pre[u] = i;
 87             }
 88     }
 89     bool solve(int s) {
 90         memset(vis, 0, sizeof(vis));
 91         if(dfs(s) != n) return false;
 92         memset(removed, 0, sizeof(removed));
 93         memset(cid, 0, sizeof(cid));
 94         for(int u = 0; u < n; u++) update(u);
 95         pre[s] = s; iw[s] = 0;
 96         ans = max_cid = 0;
 97         for(;;) {
 98             bool have_cycle = false;
 99             for(int u = 0; u < n; u++) if(u != s && !removed[u] && cycle(u)) {
100                     have_cycle = true;
101                     int v = u;
102                     do {
103                         if(v != u) removed[v] = 1;
104                         ans += iw[v];
105                         for(int i = 0; i < n; i++) if(cid[i] != cid[u] && !removed[i]) {
106                             if(w[i][v] < INF) w[i][u] = min(w[i][u], w[i][v]-iw[v]);
107                             w[u][i] = min(w[u][i], w[v][i]);
108                             if(pre[i] == v) pre[i] = u;
109                         }
110                         v = pre[v];
111                     } while(v != u);
112                     update(u); break;
113                 }
114             if(!have_cycle) break;
115         }
116         for(int i = 0; i < n; i++)
117             if(!removed[i]) ans += iw[i];
118         return true;
119     }
120 } solver;
121 int a[maxn], id[maxn];
122 
123 int main(){
124     int N, M;
125     wh(sf("%d%d", &N, &M) != EOF){
126         if(!N && !M) break;
127         id[1] = 0; int all = 0;
128         REP(i, N){
129             sf("%d", &a[i]);
130             id[i + 1] = id[i] + a[i] + 1;
131             all += (a[i] + 1);
132         }
133         solver.init(all + 1);
134         REP(i, N){
135             solver.AddEdge(0, id[i] + 1, 0);
136             REP(j, a[i])
137                 solver.AddEdge(id[i] + j + 1, id[i] + j, 0);
138         }
139         FOR(i, M){
140             int c, l1, d, l2, cost; sf("%d%d%d%d%d", &c, &l1, &d, &l2, &cost);
141             solver.AddEdge(id[c] + l1 + 1, id[d] + l2 + 1, cost);
142         }
143         if(solver.solve(0))
144             pf("%d\n", solver.ans);
145         else
146             pf("-1\n");
147     }
148 }