最大權閉合圖。題意我不是很理解,如果按英語原句翻譯的話。我覺得輸入的正值是虧損,負值是利潤。這個可以參考例樣,例樣解僱了4、5,獲得了2的利潤。對於第二問如何建圖,可以參考其他博客。這個需要畫到紙上慢慢理解,不好解釋。
對於第一問,仔細分析可以發現:
1.dfs過程不會訪問到匯點,否則最大流過程沒有完成。
2.如果一條路徑上的員工收支平衡,那麼這條路徑上的所有員工都不會被訪問到。
3.用標記數組標記已訪問過的點之後,不需要擔心最大流建圖中反向邊的影響。
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <memory.h>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iostream>
#include <sstream>
#define ll __int64
using namespace std;
#define typec ll
#define inf 1000000000
#define E 65010
#define V 5005
struct Dinic {
struct edge { int x, y, nxt; typec c; } bf[2 * E];
int ne, head[V], cur[V], ps[V], dep[V];
void init(int n) {
memset(head, 0, (n+1) * sizeof(int));
ne = 2;
}
void addedge(int x, int y, typec c) {
bf[ne].x = x; bf[ne].y = y; bf[ne].c = c;
bf[ne].nxt = head[x]; head[x] = ne++;
bf[ne].x = y; bf[ne].y = x; bf[ne].c = 0;
bf[ne].nxt = head[y]; head[y] = ne++;
}
typec flow(int n, int s, int t) {
typec tr, res = 0;
int i, j, k, f, r, top;
while (true) {
memset(dep, -1, (n+1) * sizeof(int));
for (f = dep[ps[0] = s] = 0, r = 1; f != r; )
for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
if (bf[j].c && -1 == dep[k = bf[j].y]) {
dep[k] = dep[i] + 1; ps[r++] = k;
if (k == t) { f = r; break; }
}
if (-1 == dep[t]) break;
memcpy(cur, head, (n+1) * sizeof(int));
for (i = s, top = 0; ; ) {
if (i == t) {
for (k = 0, tr = inf; k < top; ++k)
if (bf[ps[k]].c < tr)
tr = bf[ps[f = k]].c;
for (k = 0; k < top; ++k)
bf[ps[k]].c -= tr, bf[ps[k]^1].c += tr;
res += tr; i = bf[ps[top = f]].x;
}
for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
if(bf[j].c && dep[i]+1 == dep[bf[j].y]) break;
if(cur[i]) {
ps[top++] = cur[i];
i = bf[cur[i]].y;
} else {
if(0 == top) break;
dep[i] = -1; i = bf[ps[--top]].x;
}
}
}
return res;
}
} D;
bool vis[5005];
int dfs(int x) {
int ret = 0;
vis[x] = 1;
for (int p = D.head[x]; p != 0; p = D.bf[p].nxt) {
if (!vis[D.bf[p].y] && D.bf[p].c > 0) {
ret++;
ret += dfs(D.bf[p].y);
}
}
return ret;
}
int main() {
int n, m, x, y;
ll v, sum;
while (scanf("%d %d", &n, &m) != EOF) {
D.init(n + 2);
sum = 0;
for (int i = 1; i <= n; i++) {
scanf("%I64d", &v);
if (v > 0) D.addedge(0, i, v), sum += v;
else if (v < 0) D.addedge(i, n+1, -v);
}
for (int i = 0; i < m; i++) {
scanf("%d %d", &x, &y);
D.addedge(x, y, inf);
}
sum -= D.flow(n + 2, 0, n + 1);
memset(vis, 0, n + 2);
printf("%d %I64d\n", dfs(0), sum);
}
return 0;
}