标签:val incr different lines star mis like tianjin hdu
2 5 7 3 3 3 0 3 1 0 0 4 5 1 3 3 2 3 4 2 4 3 1 5 6 4 5 3 1 4 4 3 4 2 6 7 -1 -1 0 1 0 2 1 0 1 1 2 3 1 2 1 2 3 6 4 5 5 5 6 3 1 4 6 2 5 5 3 6 4
9 6
题意:有N个岛屿 M条无向路 每一个路有一最大同意的客流量。求从最西的那个岛屿最多能运用多少乘客到最东的那个岛屿。
题解:最大流,起点为最左的点,终点为最右的点。
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#define N 100020
#define ll long long
using namespace std;
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge {
int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
int n,m;
void init() {
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0) {
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end) {
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while(front != rear) {
int u = Q[front++];
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}
int S[MAXN];
int sap(int start,int end,int n) {
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top = 0;
int u = start;
int ans = 0;
while(dep[start] < n) {
if(u == end) {
int Min = INF;
int inser;
for(int i = 0; i < top; i++)
if(Min > edge[S[i]].cap - edge[S[i]].flow) {
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
for(int i = 0; i < top; i++) {
edge[S[i]].flow += Min;
edge[S[i]^1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top]^1].to;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1; i = edge[i].next) {
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) {
flag = true;
cur[u] = i;
break;
}
}
if(flag) {
S[top++] = cur[u];
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if(u != start)u = edge[S[--top]^1].to;
}
return ans;
}
int main() {
//freopen("test.in","r",stdin);
int t;
cin>>t;
while(t--) {
scanf("%d%d",&n,&m);
int s,t;
int xmin=INF,xmax=-INF;
int x,y,c;
for(int i=1; i<=n; i++) {
scanf("%d%d",&x,&y);
if(x<xmin) {
xmin=x;
s=i;
}
if(x>xmax) {
xmax=x;
t=i;
}
}
init();
for(int i=1; i<=m; i++) {
scanf("%d%d%d",&x,&y,&c);
addedge(x,y,c,c);
}
printf("%d\n",sap(s,t,n));
}
return 0;
}
Hdu 4280 Island Transport(最大流)
标签:val incr different lines star mis like tianjin hdu
原文地址:http://www.cnblogs.com/yxysuanfa/p/6999271.html