poj 3249 Test for Job (拓扑排序)

题目链接：http://poj.org/problem?id=3249

给出每个点的价值以及有向边，然后求所有路径中，价值和最大的路径，问最大价值为多少。

路径要求从入度为0的点出发，出度为0的点停止，价值可能为负值。

直接用拓扑排序，并不断向后累加每个点在价值，最后找出度为0的点的价值就可以了。

#include <cstdio>
#include <queue>
#include <vector>

using namespace std;
const int maxn = 100005;
const int oo = 1 << 31;

vector<int> G[maxn];

int weight[maxn];
int sumw[maxn];
int indeg[maxn];
int outdeg[maxn];

void topo(int n);

int main() {
int n, m, from, to;
//    freopen("1.in", "r", stdin);
while (scanf("%d%d", &n, &m) != EOF) {
for (int i = 1; i <= n; i ++) {
scanf("%d", weight + i);
}
for (int i = 0; i < m; i ++) {
scanf("%d%d", &from, &to);
G[from].push_back(to);
outdeg[from] ++;
indeg[to] ++;
}
topo(n);
for (int i = 1; i <= n; i ++) {
G[i].clear();
indeg[i] = outdeg[i] = 0;
}
}
return 0;
}

void topo(int n) {
queue<int> que;
for (int i = 1; i <= n; i ++) {
if(indeg[i] == 0) {
que.push(i);
sumw[i] =weight[i];
}
else {
sumw[i] = -oo;
}
}
while (!que.empty()) {
int cur = que.front();
que.pop();
for (int i = 0, qs = G[cur].size(); i < qs; i ++) {
int to = G[cur][i];
sumw[to] = max(sumw[to], sumw[cur] + weight[to]);
indeg[to] --;
if(indeg[to] == 0) {
que.push(to);
}
}
}
int ans = -oo;
for (int i = 1; i <= n; i ++) {
if(outdeg[i] == 0) {
if(ans < sumw[i]) {
ans = sumw[i];
}
}
sumw[i] = -oo;
}
printf("%d\n", ans);
}