題干：

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

2 2 14 21 0 1 1 0

Sample Output

35

Hint

32-bit signed integer type is capable of doing all arithmetic.

題目大意：

Flymouse從武漢大學ACM集訓隊退役后，做起了志愿者，在圣誕節來臨時，Flymouse要打扮成圣誕老人給集訓隊員發放禮物。集訓隊員住在校園宿舍的不同寢室，為了節省體力，Flymouse決定從某一個寢室出發，沿著有向路一個接一個的訪問寢室并順便發放禮物，直至能到達的所有寢室走遍為止。對于每一個寢室他可以經過無數次但是只能進入一次，進入房間會得到一個數值（數值可正可負），他想知道他能獲得最大的數值和。

解題報告：

? ?因為每一個scc中可以無限次走過，但是不一定進入房間（也就是獲得對應點的權值），所以我們只需要記錄每一個scc中有正貢獻的點。然后對縮點完的新圖的每個起點進行DP求最長路就可以了。

注意代碼姿勢非常重要，比如你如果最后建新圖的邊是遍歷每一條邊，就必須要新開一個TOT，否則會無限循環。當然，如果你不是遍歷到tot，而是遍歷到m，那就不存在這個問題了。還有就是set去重時的姿勢，別寫錯了。還有初始化，別落下tot和TOT。。其他的就沒啥了，挺簡單的一道題。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 3e5 + 5; int n,m; set<int> ss[MAX]; struct Edge {int to,fr;int ne; } e[MAX*10],E[MAX*10]; int head[MAX],DFN[MAX],LOW[MAX],stk[MAX],vis[MAX],col[MAX],clk,index,scc; int HEAD[MAX]; int val[MAX],VAL[MAX],IN[MAX],tot,TOT,TMP[MAX]; int DIS[MAX],VIS[MAX]; void add(int u,int v) {//放棄傳head數組進來了，因為還得重新弄個tot。 所以還不如直接寫兩個函數 。以后公用函數時多看看是否有重名的全局變量！！這一般都是坑 e[++tot].fr = u;e[tot].to = v;e[tot].ne = head[u];head[u] = tot; } void ADD(int u,int v) {//放棄傳head數組進來了，因為還得重新弄個tot。 所以還不如直接寫兩個函數 。以后公用函數時多看看是否有重名的全局變量！！這一般都是坑 E[++TOT].fr = u;E[TOT].to = v;E[TOT].ne = HEAD[u];HEAD[u] = TOT; } void init() {for(int i = 1; i<=n; i++) {head[i] = HEAD[i] = -1;//其實不用HEAD的初始化 下面有了 DFN[i] = LOW[i] = col[i] = vis[i] = 0;DIS[i] = VAL[i] = VIS[i] = IN[i] = TMP[i] = 0;}clk=index=scc=0;tot=TOT=0; } void Tarjan(int x) {DFN[x] = LOW[x] = ++clk;stk[++index] = x;vis[x] = 1;for(int i = head[x]; ~i; i = e[i].ne) {int v = e[i].to;if(DFN[v] == 0) {Tarjan(v);LOW[x] = min(LOW[x],LOW[v]);}else if(vis[v]) LOW[x] = min(LOW[x],DFN[v]);}if(DFN[x] == LOW[x]) {scc++;while(1) {int tmp = stk[index];index--;vis[tmp]=0;col[tmp] = scc;if(val[tmp] > 0) VAL[scc] += val[tmp];if(tmp == x) break;}} } void dfs(int cur,int rt) {if(VIS[cur]) return;VIS[cur] = 1;for(int i = HEAD[cur]; ~i; i = E[i].ne) {int v = E[i].to;if(v == rt) continue;dfs(v,cur);DIS[cur] = max(DIS[cur],DIS[v]);}DIS[cur] += VAL[cur]; } int main() {while(~scanf("%d%d",&n,&m)) {init();for(int i = 1; i<=n; i++) scanf("%d",val+i);for(int u,v,i = 1; i<=m; i++) {scanf("%d%d",&u,&v);u++,v++;add(u,v);}for(int i = 1; i<=n; i++) {if(DFN[i] == 0) Tarjan(i); }for(int i = 1; i<=scc; i++) HEAD[i] = -1,ss[i].clear();for(int i = 1; i<=tot; i++) {int u = e[i].fr;int v = e[i].to;u = col[u],v = col[v];if(u == v) continue; if(ss[u].find(v) == ss[u].end()) {ADD(u,v),IN[v]++;ss[u].insert(v);}}int cnt = 0;for(int i = 1; i<=scc; i++) {if(IN[i] == 0) TMP[++cnt] = i;}for(int i = 1; i<=cnt; i++) {dfs(TMP[i],-1);}int ans = 0;for(int i = 1; i<=cnt; i++) {ans = max(ans,DIS[TMP[i]]);}printf("%d\n",ans);}return 0 ; }

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