【连通图|强连通分量+dfs】POJ-3160 Father Christmas flymouse

Father Christmas flymouse
Time Limit: 1000MS Memory Limit: 131072K

Description

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.

Source
POJ Monthly–2006.12.31, Sempr

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题意： 给出一张有向图，图上各点有权值，权值可能为负，选择图上一个顶点出发，每个点的权值可以选择加或者不加1，求可能得到的最大权值和。

思路： 首先对于一个强连通分量，从任意一个点进入都可以到达分量中所有的点，因此先用Tarjan算法进行缩点。由于负权点根本不用加上权值，因此负权点权值设为02。

然后尝试对每个入度为0的点进行dfs，求出最大值。
代码如下：

/* * ID: j.sure.1 * PROG: * LANG: C++ */#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <climits>#include <iostream>#define PB push_back#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;const double eps = 1e-8;/****************************************/const int N = 3e4 + 5, M = 15e4 + 5;struct Edge {    int v, next;    Edge(){}    Edge(int _v, int _next):        v(_v), next(_next){}}e[M];int tot, deep, scc_cnt, n, m;int head[N], dfn[N], scc_id[N], w[N];int line[M][2], in[N], sum[N];stack <int> s;vector <int> ee[N];void init(){    tot = deep = scc_cnt = 0;    memset(head, -1, sizeof(head));    memset(dfn, 0, sizeof(dfn));    memset(scc_id, 0, sizeof(scc_id));    memset(in, 0, sizeof(in));    memset(sum, 0, sizeof(sum));}void add(int u, int v){    e[tot] = Edge(v, head[u]);    head[u] = tot++;}int dfs(int u){    int lowu = dfn[u] = ++deep;    s.push(u);    for(int i = head[u]; ~i; i = e[i].next) {        int v = e[i].v;        if(!dfn[v]) {            int lowv = dfs(v);            lowu = min(lowu, lowv);        }        else if(!scc_id[v]) {            lowu = min(lowu, dfn[v]);        }    }    if(lowu == dfn[u]) {        scc_cnt++;        while(1) {            int x = s.top(); s.pop();            scc_id[x] = scc_cnt;            sum[scc_cnt] += w[x];            if(x == u) break;        }    }    return lowu;}void tarjan(){    for(int i = 0; i < n; i++) {        if(!dfn[i]) dfs(i);    }}int dfs2(int u, int tmp){    int sz = ee[u].size();    int maxi = tmp;    for(int i = 0; i < sz; i++) {        int v = ee[u][i];        maxi = max(maxi, dfs2(v, tmp + sum[v]));    }    return maxi;}int solve(){    int ans = 0;    for(int i = 1; i <= scc_cnt; i++) ee[i].clear();    for(int i = 0; i < m; i++) {        int u = scc_id[line[i][0]], v = scc_id[line[i][1]];        if(u != v) {            in[v]++;            ee[u].PB(v);        }    }    for(int i = 1; i <= scc_cnt; i++) {        if(!in[i]) {            ans = max(ans, dfs2(i, sum[i]));        }    }    return ans;}int main(){#ifdef J_Sure    freopen("000.in", "r", stdin);    //freopen("999.out", "w", stdout);#endif    while(~scanf("%d%d", &n, &m)) {        init();        for(int i = 0; i < n; i++) {            scanf("%d", &w[i]);            if(w[i] < 0) w[i] = 0;        }        int u, v;        for(int i = 0; i < m; i++) {            scanf("%d%d", &u, &v);            add(u, v);            line[i][0] = u; line[i][1] = v;        }        tarjan();        int ans = solve();        printf("%d\n", ans);    }    return 0;}

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1. 题意描述不清楚，假如整张图都是负权，答案是0。 ↩
2. 不得不怀疑出题人是否脑残。 ↩