POJ 3160 Father Christmas flymouse

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mob60475703f08d 2014-11-20 20:50:00

文章标签 强连通最长路 #include #define ios 文章分类 后端开发

Father Christmas flymouse

Time Limit: 1000ms

Memory Limit: 131072KB

This problem will be judged on PKU. Original ID: 3160
64-bit integer IO format: %lld Java class name: Main

After retirement as contestant from WHU ACM Team, flymouse volunteered to do the odds and ends such as cleaning out the computer lab for training as extension of his contribution to the team. When Christmas came, flymouse played Father Christmas to give gifts to the team members. The team members lived in distinct rooms in different buildings on the campus. To save vigor, flymouse decided to choose only one of those rooms as the place to start his journey and follow directed paths to visit one room after another and give out gifts en passant until he could reach no more unvisited rooms.

During the days on the team, flymouse left different impressions on his teammates at the time. Some of them, like LiZhiXu, with whom flymouse shared a lot of candies, would surely sing flymouse’s deeds of generosity, while the others, like snoopy, would never let flymouse off for his idleness. flymouse was able to use some kind of comfort index to quantitize whether better or worse he would feel after hearing the words from the gift recipients (positive for better and negative for worse). When arriving at a room, he chould choose to enter and give out a gift and hear the words from the recipient, or bypass the room in silence. He could arrive at a room more than once but never enter it a second time. He wanted to maximize the the sum of comfort indices accumulated along his journey.

Input

The input contains several test cases. Each test cases start with two integers N and M not exceeding 30 000 and 150 000 respectively on the first line, meaning that there were N team members living in N distinct rooms and M direct paths. On the next N lines there are N integers, one on each line, the i-th of which gives the comfort index of the words of the team member in the i-th room. Then follow M lines, each containing two integers i and j indicating a directed path from the i-th room to the j-th one. Process to end of file.

Output

For each test case, output one line with only the maximized sum of accumulated comfort indices.

Sample Input

Sample Output

Source

POJ Monthly--2006.12.31

解题：强连通缩点+最长路

POJ 3160 Father Christmas flymouse_#define

POJ 3160 Father Christmas flymouse_最长路_02

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <climits>
  7 #include <vector>
  8 #include <queue>
  9 #include <cstdlib>
 10 #include <string>
 11 #include <set>
 12 #include <stack>
 13 #define LL long long
 14 #define pii pair<int,int>
 15 #define INF 0x3f3f3f3f
 16 using namespace std;
 17 const int maxn = 30010;
 18 struct arc{
 19     int to,next;
 20     arc(int x = 0,int y = -1){
 21         to = x;
 22         next = y;
 23     }
 24 };
 25 arc e[maxn*10];
 26 vector<int>g[maxn];
 27 int tot,head[maxn],d[maxn],dfn[maxn],low[maxn],belong[maxn];
 28 int hv[maxn],thv[maxn],scc,idx;
 29 int my[maxn],in[maxn],out[maxn],top,n,m;
 30 bool instack[maxn],inq[maxn];
 31 void add(int u,int v){
 32     e[tot] = arc(v,head[u]);
 33     head[u] = tot++;
 34 }
 35 void tarjan(int u){
 36     dfn[u] = low[u] = ++idx;
 37     my[top++] = u;
 38     instack[u] = true;
 39     for(int i = head[u]; ~i; i = e[i].next){
 40         if(!dfn[e[i].to]){
 41             tarjan(e[i].to);
 42             low[u] = min(low[u],low[e[i].to]);
 43         }else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]);
 44     }
 45     if(low[u] == dfn[u]){
 46         int v;
 47         scc++;
 48         do{
 49             v = my[--top];
 50             instack[v] = false;
 51             belong[v] = scc;
 52         }while(v != u);
 53     }
 54 }
 55 void init(){
 56     for(int i = 0; i <= n; ++i){
 57         head[i] = -1;
 58         dfn[i] = low[i] = 0;
 59         out[i] = belong[i] = 0;
 60         inq[i] = instack[i] = false;
 61         g[i].clear();
 62         hv[i] = thv[i] = 0;
 63         in[i] = d[i] = 0;
 64     }
 65     tot = idx = scc = 0;
 66 }
 67 void spfa(){
 68     queue<int>q;
 69     q.push(0);
 70     while(!q.empty()){
 71         int u = q.front();
 72         q.pop();
 73         inq[u] = false;
 74         for(int i = g[u].size()-1; i >= 0; --i){
 75             if(d[g[u][i]] < d[u] + thv[g[u][i]]){
 76                 d[g[u][i]] = d[u] + thv[g[u][i]];
 77                 if(!inq[g[u][i]]){
 78                     inq[g[u][i]] = true;
 79                     q.push(g[u][i]);
 80                 }
 81             }
 82         }
 83     }
 84 }
 85 int main() {
 86     int u,v,w;
 87     while(~scanf("%d %d",&n,&m)){
 88         init();
 89         for(int i = 0; i < n; ++i)
 90             scanf("%d",hv+i);
 91         for(int i = 0; i < m; ++i){
 92             scanf("%d %d",&u,&v);
 93             add(u,v);
 94         }
 95         for(int i = 0; i < n; ++i)
 96             if(!dfn[i]) tarjan(i);
 97         for(int i = 0; i < n; ++i)
 98             if(hv[i] > 0) thv[belong[i]] += hv[i];
 99         for(int i = 0; i < n; ++i){
100             for(int j = head[i]; ~j; j = e[j].next){
101                 if(belong[i] == belong[e[j].to]) continue;
102                 g[belong[i]].push_back(belong[e[j].to]);
103                 in[belong[e[j].to]]++;
104                 out[belong[i]]++;
105             }
106         }
107         for(int i = 1; i <= scc; ++i)
108             if(!in[i]) g[0].push_back(i);
109         int ans = 0;
110         spfa();
111         for(int i = 1; i <= scc; ++i)
112             if(!out[i]) ans = max(ans,d[i]);
113         printf("%d\n",ans);
114     }
115     return 0;
116 }