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poj 1273 Drainage Ditches | hdoj 1532

2012-03-01 13:14 337 查看
类型:最大流

经典题目 本题同hdoj 1532

dinic算法实现 [可做模板]

!!! ne 需要从2开始,正向边的编号为偶数,反向边的编号为奇数

如果从1开始,那么当更新编号为3的边时,其反向边的编号为4,而此时更新的是编号为2的边,出现错误

// hdoj 1532 | poj 1273
// poj 0ms hdoj 15ms
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const int inf = 1 << 30;
const int MAXN = 2000;

struct node {
int c,u,v,next;
}edge[MAXN];
int ne, head[MAXN];
int cur[MAXN], ps[MAXN], dep[MAXN];

void addedge(int u, int v,int w) {
edge[ne].u = u;
edge[ne].v = v;
edge[ne].c = w;
edge[ne].next = head[u];
head[u] = ne ++;
edge[ne].u = v;
edge[ne].v = u;
edge[ne].c = 0;
edge[ne].next = head[v];
head[v] = ne ++;
}

int dinic(int s, int t) {                       //  dinic
int tr, res = 0;
int i, j, k, f, r, top;
while(1) {
memset(dep, -1, sizeof(dep));
for(f = dep[ps[0] = s] = 0, r = 1; f != r;)
for(i = ps[f++], j = head[i]; j != -1; j = edge[j].next)
if(edge[j].c && dep[k = edge[j].v] == -1) {
dep[k] = dep[i] + 1;
ps[r++] = k;
if(k == t) {
f = r; break;
}
}
if(dep[t] == -1) break;
memcpy(cur, head, sizeof(cur));
i = s, top = 0;
while(1) {
if(i == t) {
for(tr = inf, k = 0; k < top; k++)
if(edge[ps[k]].c < tr)
tr = edge[ps[f = k]].c;
for(k = 0; k < top; k++) {
edge[ps[k]].c -= tr;
edge[ps[k]^1].c += tr;
}
i = edge[ps[top=f]].u;
res += tr;
}
for(j = cur[i]; cur[i] != -1; j = cur[i] = edge[cur[i]].next) {
if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
}
if(cur[i] != -1) {
ps[top++] = cur[i];
i = edge[cur[i]].v;
}
else {
if(top == 0) break;
dep[i] = -1;
i = edge[ps[--top]].u;
}
}
}
return res;
}

int main() {
int cas, i, j, M, N, a, b, c;

while(scanf("%d %d", &N, &M) == 2) {
ne = 2;
memset(head,-1,sizeof(head));
for(i = 0; i < N; ++i) {
scanf("%d %d %d", &a, &b, &c);
addedge(a, b, c);
}
cout<<dinic(1, M)<<endl;
}
return 0;
}
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