hdu 4280 Island Transport isap果题

题意：给出n个二维坐标系上的点（其实就是n个点），最左的是起点，最右的是终点，且各只有一个，问从起点到终点的最大流。

建边直接上模板即可

#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <string>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8#define asd puts("sdfsdfsdfsdfsdfsdf");typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 100010;struct node{    int v, w, nxt;}e[N<<2];int head[N];int cur[N];int gap[N];int s[N], top;int dep[N];int st, ed;int n, m, cnt;queue <int> q;void init(){    cnt = 0;    memset( head, -1, sizeof( head ) );}void add( int u, int v, int w ){    e[cnt].v = v;    e[cnt].w = w;    e[cnt].nxt = head[u];    head[u] = cnt++;    e[cnt].v = u;    e[cnt].w = w;    e[cnt].nxt = head[v];    head[v] = cnt++;}void rev_bfs(){    memset( gap, 0, sizeof( gap ) );    memset( dep, -1, sizeof( dep ) );    while( !q.empty() ) q.pop();    q.push(ed);    dep[ed] = 0;    gap[0] = 1;    while( !q.empty() ) {        int u = q.front();        q.pop();        for( int i = head[u]; ~i; i = e[i].nxt ) {            int v = e[i].v;            if( ~dep[v] )                continue;            dep[v] = dep[u] + 1;            gap[dep[v]]++;            q.push(v);        }    }}int isap(){    int flow = 0, u = st, i;    memcpy( cur, head, sizeof head );    rev_bfs();    top = 0;    //int nv = ed + 1;    while( dep[st] < n ) {        if( u == ed ) {            int tmp, minn = inf;            for( i = 0; i < top; ++i ) {                if( minn > e[s[i]].w ) {                    minn = e[s[i]].w;                    tmp = i;                }            }            for( i = 0; i < top; ++i ) {                e[s[i]].w -= minn;                e[s[i]^1].w += minn;            }            flow += minn;            top = tmp;            u = e[s[top]^1].v;        }        for( i = cur[u]; ~i; i = e[i].nxt ) {            if( e[i].w > 0 && dep[u] == dep[e[i].v] + 1 ) {                cur[u] = i;                break;            }        }        if( ~i ) {            s[top++] = cur[u];            u = e[i].v;        }        else {            if( 0 == (--gap[dep[u]]) )                break;            int minn = n;            for( i = head[u]; ~i; i = e[i].nxt ) {                int v = e[i].v;                if( e[i].w > 0 && minn > dep[v] ) {                    minn = dep[v];                    cur[u] = i;                }            }            dep[u] = minn + 1;            gap[dep[u]]++;            if( u != st ) {                u = e[ s[--top]^1 ].v;            }        }    }    return flow;}int main(){    int tot;    scanf("%d", &tot);    while( tot-- ) {        int t1 = inf, t2 = -inf;    ///t1 for st, t2 for ed        init();        scanf("%d%d", &n, &m);        int x, y;        for( int i = 1; i <= n; ++i ) {            scanf("%d%d", &x, &y);            if( x < t1 ) {                st = i;                t1 = x;            }            if( x > t2 ) {                t2 = x;                ed = i;            }        }        int u, v, w;        while( m-- ) {            scanf("%d%d%d", &u, &v, &w);            add( u, v, w );        }        int ans = isap();        printf("%d\n", ans);    }    return 0;}