Sicily 1827/1947. Snipers

1827. Snipers

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Snipers play an important role in wars. They kill enemy soldiers who are far away. A small team of snipers can stop hundreds of infantrymen. When in war, a commander must set snipers in the most valuable positions.

       You are the commander today. The enemy is approaching. You must send out your snipers to stop them. There are N+1 positions, which are identified by 0, 1… N. The enemy soldiers set out at position 0. Their destination is position N. They can go from one position to another if the two positions are “connected”. You are in position N and you don’t want any enemy soldiers get to you. The snipers can only be set in position 1, 2… N-1.

       For position i (0 < i < N), no one can pass by if there are no less than Ki (0 < Ki< 100) snipers in it. That means no enemy soldiers can come to the position and then go ahead to another. They will be shot before they arrive. Please note that position 0 and N are NOT connected.

Input

The first line contains an integer indicating the number of test cases.

For each test case, two integers N and M (0 < N < 50, 0 < M < 200) are in the first line. The next line has N-1 integers. They are K₁, K₂… K_N-1. Following are M lines. Each contains a pair of integers A and B (0 < A, B < N), which means that position A and B are “connected”. There is NO bland line between two cases.

Output

For each test case, output the less snipers needed, one case per line. There is NO bland line between two cases.

Sample Input

14 52 3 40 10 31 22 43 4

Sample Output

6

// Problem#: 1827// Submission#: 3589660// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <string.h>const int MAXN = 105;const int MAXNN = MAXN / 2;int graph[MAXN][MAXN];int node[MAXNN];int n, c[MAXN + 1][MAXN + 1], s, t;int f[MAXN + 1][MAXN + 1];int pre[MAXN + 1], tag[MAXN + 1];int find() {    int queue[MAXN + 10], head, tail;    int x, loop;    for (loop = 0; loop <= n; loop++) tag[loop] = 0;    queue[0] = s;    head = 0;    tail = 1;    tag[s] = 1;    while (head != tail) {        x = queue[head++];        if (head == MAXN + 10) head = 0;        for (loop = 0; loop <= n; loop++) if (tag[loop] == 0) {            if (f[x][loop] < c[x][loop]) {                pre[loop] = x;                tag[loop] = 1;                if (loop == t) return 1;                queue[tail++] = loop;                if (tail == MAXN + 10) tail = 0;            } else if (f[loop][x] > 0) {                pre[loop] = x;                tag[loop] = -1;                if (loop == t) return 1;                queue[tail++] = loop;                if (tail == MAXN + 10) tail = 0;            }        }    }    return 0;}void adjust() {    int min = 2000000000, loop;    for (loop = t; loop != s; loop = pre[loop]) {        if (tag[loop] > 0) {            if (c[pre[loop]][loop] - f[pre[loop]][loop] < min)                min = c[pre[loop]][loop] - f[pre[loop]][loop];        } else {            if (f[loop][pre[loop]] < min)                min = f[loop][pre[loop]];        }    }    for (loop = t; loop != s; loop = pre[loop]) {        if (tag[loop] > 0) {            f[pre[loop]][loop] += min;        } else {            if (f[loop][pre[loop]] < min)                f[loop][pre[loop]] -= min;        }    }}void solve() {    int i, j;    for (i = 0; i <= n; i++) for (j = 0; j <= n; j++) f[i][j] = 0;    while (find()) adjust();}int main() {    int testcase, nn, mm;    scanf("%d", &testcase);    while (testcase--) {        memset(graph, 0, sizeof(graph));        memset(node, 0, sizeof(node));        scanf("%d%d", &nn, &mm);        for (int i = 1; i <= nn - 1; i++) scanf("%d", node + i);        for (int i = 0; i < mm; i++) {            int a1, a2, t;            scanf("%d%d", &a1, &a2);            if (a1 > a2) {                t = a1;                a1 = a2;                a2 = t;            }            graph[a1][a2] = 1;            if (a2 != nn && a1 != 0) graph[a2][a1] = 1;        }        memset(c, 0, sizeof(c));        for (int i = 1; i <= nn - 1; i++)            c[i][i + nn] = node[i];        for (int i = 0; i <= nn; i++)            for (int j = 0; j <= nn; j++)                if (graph[i][j]) {                    int t1, t2;                    if (i == 0) {                        t1 = 0;                        t2 = j;                    } else {                        t1 = i + nn;                        t2 = j;                    }                    c[t1][t2] = 2000000000;                }                n = nn * 2 - 1;                s = 0;                t = nn;                solve();                int maxFlow = 0;                for (int i = 0; i <= n; i++)  maxFlow += f[i][nn];                printf("%d\n", maxFlow);    }    return 0;}