poj - 2455 - Secret Milking Machine（最大流+二分）

题意：N个点，P条边（每条边有边长）的无向图，求结点1到结点N的T条路径（每条边只能用一次）中，最长边的最小值(2 <= N <= 200, 1 <= P <= 40,000)。

题目链接：http://poj.org/problem?id=2455

——>>二分答案。。。

建图：设超级源S = 0，S到1连一边容量为T的有向边，连长 <= 答案的双向加入，容量为1，最后N到超级汇T连一条容量为无穷大的有向边。

G++TLE。。。

C++375MS。。。

另外，数组范围要大一点点的样子。。。

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn = 200 + 10;const int maxm = 400000 + 10;const int INF = 0x3f3f3f3f;int N, P, T;int a[maxm], b[maxm], w[maxm];int head[maxn], v[maxm<<1], nxt[maxm<<1], cap[maxm<<1], flow[maxm<<1], h[maxn], cur[maxn], ecnt;struct Dinic {    int s, t;    Dinic() {        memset(head, -1, sizeof(head));        ecnt = 0;    }    void addEdge(int uu, int vv, int ca) {        v[ecnt] = vv; cap[ecnt] = ca; flow[ecnt] = 0; nxt[ecnt] = head[uu]; head[uu] = ecnt++;        v[ecnt] = uu; cap[ecnt] = 0; flow[ecnt] = 0; nxt[ecnt] = head[vv]; head[vv] = ecnt++;    }    bool bfs() {        memset(h, -1, sizeof(h));        h[s] = 0;        queue<int> qu;        qu.push(s);        while(!qu.empty()) {            int u = qu.front(); qu.pop();            for(int e = head[u]; e != -1; e = nxt[e])                if(h[v[e]] == -1 && cap[e] > flow[e]) {                    h[v[e]] = h[u] + 1;                    qu.push(v[e]);                }        }        return h[t] != -1;    }    int dfs(int u, int a) {        if(u == t || !a) return a;        int f, Flow = 0;        for(int e = cur[u]; e != -1; e = nxt[e]) {            cur[u] = e;            if(h[v[e]] == h[u] + 1 && (f = dfs(v[e], min(a, cap[e]-flow[e]))) > 0) {                flow[e] += f;                flow[e^1] -= f;                Flow += f;                a -= f;                if(!a) break;            }        }        return Flow;    }    int maxflow(int s, int t) {        this->s = s;        this->t = t;        int Flow = 0;        while(bfs()) {            memcpy(cur, head, sizeof(head));            Flow += dfs(s, INF);        }        return Flow;    }};bool isok(int len) {    Dinic din;    din.addEdge(0, 1, T);    for(int i = 0; i < P; i++) if(w[i] <= len) {        din.addEdge(a[i], b[i], 1);        din.addEdge(b[i], a[i], 1);    }    din.addEdge(N, N+1, INF);    return din.maxflow(0, N+1) == T;}int main(){    while(scanf("%d%d%d", &N, &P, &T) == 3) {        int L = INF, R = -1;        for(int i = 0; i < P; i++) {            scanf("%d%d%d", &a[i], &b[i], &w[i]);            if(w[i] < L) L = w[i];            if(w[i] > R) R = w[i];        }        L--;        R++;        int ret;        while(L + 1 < R) {            int M = (L + R) >> 1;            if(isok(M)) {                ret = M;                R = M;            }            else L = M;        }        printf("%d\n", ret);    }    return 0;}