poj-1698-Alice's Chance

这道题主要难点是构图，想了半天也想不懂怎么构图，后来在网上找了一下别人的题解

构图：把每个电影需要的星期数都拆开，变成一条线的权值设为1，这样每个星期中的某一天只能被一部电影占据，源点直接连接到电影，权值为需要的天数，之后将日期和汇点相连，比如有一部电影需要4周，就需要有4*7=28个点和汇点相连！

把图画出来就直接用dinic算法就行了

Dinic算法的原理与构造

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
#define N 400
const int INF = 0x3f3f3f3f;
using namespace std;
struct Edge{
int v, f;
int next;
}edge[N*N];
int n, first
, cnt, q
, F[22][8], level
;
void Init(){
cnt = 0;
memset(first, -1, sizeof(first));
}
void read(int u, int v, int f){
edge[cnt].v = v;
edge[cnt].f = f;
edge[cnt].next = first[u];
first[u] = cnt++;
edge[cnt].v = u;
edge[cnt].f = 0;
edge[cnt].next = first[v];
first[v] = cnt++;
}
bool bfs(int s, int t){
memset(level, 0, sizeof(level));
level[s] = 1;
int rear = 0, front = 0;
q[front++] = s;
while(rear < front){
int x = q[rear++];
if (x == t) return true;
for (int e = first[x]; e != -1; e = edge[e].next){
int v = edge[e].v;
int f = edge[e].f;
if (!level[v] && f){
level[v] = level[x]+1;
q[front++] = v;
}
}
}
return false;
}
int Dinic(int s, int t){
int ans = 0;
while(bfs(s, t)){
int e, x, y, back, iter = 1;
while(iter){
x = (iter==1)?s:edge[q[iter-1]].v;
if (x == t){
int minCap = INF;
for (int i = 1; i < iter; i++){
e = q[i];
if (edge[e].f < minCap){
minCap = edge[e].f;
back = i;
}
}
for (int i = 1; i < iter; i++){
e = q[i];
edge[e].f -= minCap;
edge[e^1].f += minCap;
}
ans += minCap;
iter = back;
}else{
for (e = first[x]; e != -1; e = edge[e].next){
y = edge[e].v;
if (edge[e].f && level[y] == level[x]+1){
break;
}
}
if (e != -1){
q[iter++] = e;
}
else{
level[x] = -1;
iter--;
}
}
}
}
return ans;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
#endif
int i, j, k, T, d, w, sum, flag;
scanf("%d", &T);
while(T--){
sum = 0;
flag = 0;
Init();
scanf("%d", &n);
for (i = 1; i <= n; i++){
for (j = 1; j <= 7; j++){
scanf("%d", &F[i][j]);
}
scanf("%d%d", &d, &w);
read(0, i, d);
sum += d;
flag = max(flag, w);
for (k = 0; k < w; k++){
for (j = 1; j <= 7; j++){
if (F[i][j]){
read(i, 7*k+j+n, 1);
}
}
}
}
int tmp = n+7*flag+1;
for (i = 0; i < flag; i++){
for (j = 1; j <= 7; j++){
read(7*i+n+j, tmp, 1);
}
}
int ans = Dinic(0, tmp);
if (sum == ans){
puts("Yes");
}else{
puts("No");
}
}
return 0;
}